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pslct037

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<title>Burnside's Theorem Foreplay</title></titleStmt>

<publicationStmt><distributor>BASE and Oxford Text Archive</distributor>

<idno>pslct037</idno>

<availability><p>The British Academic Spoken English (BASE) corpus was developed at the

Universities of Warwick and Reading, under the directorship of Hilary Nesi

(Centre for English Language Teacher Education, Warwick) and Paul Thompson

(Department of Applied Linguistics, Reading), with funding from BALEAP,

EURALEX, the British Academy and the Arts and Humanities Research Board. The

original recordings are held at the Universities of Warwick and Reading, and

at the Oxford Text Archive and may be consulted by bona fide researchers

upon written application to any of the holding bodies.

The BASE corpus is freely available to researchers who agree to the

following conditions:</p>

<p>1. The recordings and transcriptions should not be modified in any

way</p>

<p>2. The recordings and transcriptions should be used for research purposes

only; they should not be reproduced in teaching materials</p>

<p>3. The recordings and transcriptions should not be reproduced in full for

a wider audience/readership, although researchers are free to quote short

passages of text (up to 200 running words from any given speech event)</p>

<p>4. The corpus developers should be informed of all presentations or

publications arising from analysis of the corpus</p><p>

Researchers should acknowledge their use of the corpus using the following

form of words:

The recordings and transcriptions used in this study come from the British

Academic Spoken English (BASE) corpus, which was developed at the

Universities of Warwick and Reading under the directorship of Hilary Nesi

(Warwick) and Paul Thompson (Reading). Corpus development was assisted by

funding from the Universities of Warwick and Reading, BALEAP, EURALEX, the

British Academy and the Arts and Humanities Research Board. </p></availability>

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<date>04/12/2002</date><equipment><p>video</p></equipment>

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<u who="nm0941"> the object <pause dur="0.2"/> of <pause dur="0.4"/> the last two lectures of this course <pause dur="0.4"/> today <pause dur="0.5"/> and Friday are to send you home in a good mood <pause dur="1.1"/> # <pause dur="0.7"/> on Friday <pause dur="0.7"/> we shall <pause dur="0.6"/> reach <pause dur="0.2"/> a climax <pause dur="1.1"/> with a proof <pause dur="2.0"/><kinesic desc="writes on board" iterated="y" dur="2"/> of a theorem i've already trailered <pause dur="0.8"/> namely Burnside's <pause dur="4.1"/><kinesic desc="writes on board" iterated="y" dur="10"/> so-called <pause dur="0.7"/> P-to-the-A <pause dur="0.3"/> Q-to-the-B <pause dur="0.5"/> theorem <pause dur="2.3"/> and today <pause dur="0.6"/> we shall indulge in a little foreplay <pause dur="2.2"/> right <pause dur="1.1"/> we're going to <trunc>u</trunc> exploit some of the facts <pause dur="3.0"/><kinesic desc="writes on board" iterated="y" dur="4"/> we have stated about algebraic numbers <pause dur="7.0"/> if you have a character <pause dur="5.8"/><kinesic desc="writes on board" iterated="y" dur="5"/> of a group G <pause dur="4.8"/><kinesic desc="writes on board" iterated="y" dur="11"/> and you have an element of the group G <pause dur="5.2"/> then <pause dur="1.1"/> a character value on G divided by the degree of that character <pause dur="2.6"/> is actually an algebraic <pause dur="4.5"/><kinesic desc="writes on board" iterated="y" dur="3"/> number <pause dur="1.1"/> and # <pause dur="4.2"/><kinesic desc="writes on board" iterated="y" dur="3"/> it satisfies <pause dur="18.0"/><kinesic desc="writes on board" iterated="y" dur="15"/> that's the easy bit <pause dur="1.4"/> the <pause dur="0.9"/> important bit is that # <pause dur="2.1"/><kinesic desc="writes on board" iterated="y" dur="23"/> if <pause dur="1.6"/> it's <pause dur="1.4"/> an algebraic integer <pause dur="2.6"/> then <pause dur="1.3"/> <vocal desc="clears throat" iterated="n"/> <pause dur="4.2"/> the absolute value <pause dur="1.2"/> of this <pause dur="1.2"/>

quotient <pause dur="0.9"/> is nought <pause dur="2.1"/> or one <pause dur="7.4"/><kinesic desc="writes on board" iterated="y" dur="3"/> well <pause dur="0.2"/> we know <pause dur="0.4"/><vocal desc="clears throat" iterated="n"/><pause dur="1.1"/> # <pause dur="0.3"/> let's <pause dur="0.9"/><kinesic desc="writes on board" iterated="y" dur="5"/> call that thing alpha to save having to keep writing it out <pause dur="3.8"/> that's certainly a complex number <pause dur="0.4"/> it's <pause dur="0.7"/> clearly an algebraic integer because <pause dur="0.6"/> we know that the character values or sums of roots of unity <pause dur="0.4"/> and i'm just dividing by <pause dur="0.8"/> an integer <pause dur="1.4"/><vocal desc="clears throat" iterated="n"/> # <pause dur="0.9"/> so <pause dur="0.2"/> this is <pause dur="0.2"/> equal<kinesic desc="writes on board" iterated="y" dur="2"/> to <pause dur="0.8"/> yes <pause dur="0.4"/> then <pause dur="1.0"/><vocal desc="clears throat" iterated="n"/><pause dur="1.1"/> # <pause dur="0.5"/> chi-of-G <pause dur="2.4"/><kinesic desc="writes on board" iterated="y" dur="17"/> is a sum <pause dur="0.9"/> of roots of unity <pause dur="13.5"/> # <pause dur="2.8"/> therefore <pause dur="12.0"/><kinesic desc="writes on board" iterated="y" dur="6"/> if i look at the absolute value of alpha <pause dur="1.3"/> this is <pause dur="1.0"/> # <pause dur="0.9"/> the absolute value <pause dur="0.2"/> of this <kinesic desc="indicates point on board" iterated="n"/> divided by D <pause dur="0.7"/> by the <pause dur="1.0"/><kinesic desc="writes on board" iterated="y" dur="8"/> triangle of # <pause dur="1.3"/> complex numbers <pause dur="7.8"/> that's less than or equal to the absolute value of the <pause dur="0.6"/> roots of unity the absolute value of a root of unity they lie on the unit circle are one <pause dur="1.5"/><kinesic desc="writes on board" iterated="y" dur="1"/> and so <pause dur="0.7"/> i've proved the <pause dur="0.3"/> inequality here <kinesic desc="indicates point on board" iterated="n"/><pause dur="1.4"/><kinesic desc="writes on board" iterated="y" dur="1"/> absolute values are never negative <pause dur="2.0"/> # <pause dur="0.3"/> suppose <pause dur="4.3"/><kinesic desc="writes on board" iterated="y" dur="22"/> now <pause dur="1.2"/> that alpha is an algebraic <pause dur="1.6"/> integer <pause dur="3.8"/> with <pause dur="0.5"/><vocal desc="clears throat" iterated="n"/><pause dur="1.8"/> # say min-<pause dur="1.6"/><vocal desc="clears throat" iterated="n"/><pause dur="1.1"/>Q <pause dur="0.2"/> alpha

the minimum polynomial that will be <pause dur="0.8"/> X to the <pause dur="1.2"/> oh we don't know what the coefficient is <kinesic desc="writes on board" iterated="y" dur="5"/> M <pause dur="0.8"/> plus A-one-<pause dur="0.2"/>X-to-the-M-<pause dur="0.5"/>minus-one <pause dur="1.3"/> plus <pause dur="0.3"/> # <pause dur="2.0"/> A-M <pause dur="2.2"/> with the A-Is <pause dur="0.4"/><vocal desc="clears throat" iterated="n"/><pause dur="1.1"/><kinesic desc="writes on board" iterated="y" dur="2"/> in Z because we're supposing <pause dur="0.5"/> that it's an algebraic integer <pause dur="7.8"/><vocal desc="clears throat" iterated="n"/><pause dur="0.7"/> assume <pause dur="2.6"/><kinesic desc="writes on board" iterated="y" dur="4"/> that alpha <pause dur="0.7"/> is strictly less than one <pause dur="1.9"/> i'll be done if i can show <pause dur="0.8"/> that this implies alpha equals nought <pause dur="6.3"/><kinesic desc="writes on board" iterated="y" dur="2"/> by one of the <pause dur="1.0"/> facts i proved <pause dur="5.4"/> well this is # <pause dur="1.2"/><kinesic desc="writes on board" iterated="y" dur="4"/> fourteen-point-eight fact one used here <pause dur="1.6"/> that's just the <pause dur="0.9"/> result that i have <pause dur="1.4"/> on the fourth line there <pause dur="0.8"/><vocal desc="sniff" iterated="n"/><pause dur="1.6"/> and by fact three <pause dur="15.2"/><kinesic desc="writes on board" iterated="y" dur="5"/> we know that the conjugates of alpha <pause dur="19.0"/><kinesic desc="writes on board" iterated="y" dur="17"/> where <pause dur="0.5"/> the omega-one up to omega-D <pause dur="0.8"/> are conjugates of the <pause dur="1.9"/> roots of unity omega-one up to omega-D <pause dur="0.5"/> and <pause dur="0.2"/> are

therefore <pause dur="0.2"/> again <pause dur="0.6"/> roots of unity <pause dur="14.8"/><kinesic desc="writes on board" iterated="y" dur="10"/> i guess i <pause dur="0.2"/><kinesic desc="writes on board" iterated="y" dur="2"/> forgot to divide out by <pause dur="0.9"/> D <pause dur="1.2"/> and this <pause dur="1.2"/><kinesic desc="writes on board" iterated="y" dur="2"/> is all divided by D <kinesic desc="indicates point on board" iterated="n"/> these are roots of unity <pause dur="0.7"/> their absolute value is <kinesic desc="writes on board" iterated="y" dur="1"/> one <pause dur="0.9"/> and so the <pause dur="0.6"/> conjugates also have absolute value <pause dur="2.8"/> equal to # <pause dur="0.6"/> or less than or equal to # <pause dur="1.1"/> # sorry the conjugates have absolute value equal to one therefore <pause dur="0.5"/> # <pause dur="0.2"/> # <kinesic desc="indicates point on board" iterated="n"/> this conjugate <pause dur="0.4"/> has absolute value at most one <pause dur="0.7"/> now <pause dur="0.4"/> # the conjugates of alpha <pause dur="0.5"/> are the roots of that polynomial <kinesic desc="indicates point on board" iterated="n"/> up there <pause dur="1.0"/> and as we know the product of the roots of a polynomial <pause dur="2.7"/> are <pause dur="0.5"/> plus or minus the constant term <pause dur="5.6"/><kinesic desc="writes on board" iterated="y" dur="4"/> if you factorize the polynomial into linear factors <pause dur="0.4"/> the constant term <pause dur="0.4"/> is the product of the roots <pause dur="10.7"/><kinesic desc="writes on board" iterated="y" dur="12"/> and that is of course an integer <pause dur="16.2"/> now <pause dur="0.8"/> the absolute value <pause dur="5.2"/><kinesic desc="writes on board" iterated="y" dur="23"/> of the

product of the conjugate <pause dur="7.7"/><vocal desc="clears throat" iterated="n"/><pause dur="0.2"/> is equal to the <pause dur="1.4"/> # <pause dur="8.4"/> things like that <pause dur="0.6"/><kinesic desc="indicates point on board" iterated="n"/> we made the assumption that alpha had absolute value <pause dur="0.2"/> strictly less than one <pause dur="0.4"/> so <pause dur="0.8"/><kinesic desc="writes on board" iterated="y" dur="1"/> because of that last line there <pause dur="0.8"/><kinesic desc="indicates point on board" iterated="n"/> the <pause dur="2.1"/><kinesic desc="writes on board" iterated="y" dur="2"/> absolute value of this is strictly less than one <pause dur="0.9"/> and <pause dur="1.3"/><kinesic desc="writes on board" iterated="y" dur="2"/> equals the <pause dur="1.0"/> absolute value of A-N <pause dur="0.9"/> and that's a natural number <pause dur="2.2"/><kinesic desc="writes on board" iterated="y" dur="1"/> now can anyone tell me a natural number <pause dur="4.0"/> whose absolute value <pause dur="1.3"/> is strictly less than one <pause dur="1.7"/> any volunteers <pause dur="1.1"/> you're on camera so you <pause dur="0.2"/> needn't be shy </u><pause dur="0.8"/> <u who="sm0942" trans="pause"> <gap reason="inaudible" extent="3 secs"/> is zero considered <pause dur="0.2"/> <gap reason="inaudible" extent="1 sec"/> </u><u who="nm0941" trans="overlap"> yeah for the purposes of <pause dur="0.5"/> # this course zero <pause dur="1.0"/> is a natural number <pause dur="1.2"/> otherwise this would not make sense <pause dur="0.4"/> therefore <pause dur="0.4"/><kinesic desc="writes on board" iterated="y" dur="2"/><vocal desc="clears throat" iterated="n"/><pause dur="0.6"/> A-N <pause dur="0.4"/> is zero <pause dur="1.3"/> and if

the constant term of an irreducible polynomial <pause dur="0.4"/> is zero <pause dur="0.4"/> then X is a factor of it but it's irreducible <pause dur="0.4"/> therefore the polynomial <pause dur="0.2"/> is <pause dur="0.4"/> X itself <pause dur="17.1"/><kinesic desc="writes on board" iterated="y" dur="15"/> and therefore <pause dur="1.2"/> # <pause dur="1.5"/> the root <pause dur="0.3"/> we have a unique root namely alpha <pause dur="1.3"/> which is zero <pause dur="10.7"/><kinesic desc="writes on board" iterated="y" dur="7"/> so we took <pause dur="1.7"/> a number alpha <pause dur="0.3"/> of the given form which was strictly less than one in absolute value <pause dur="0.4"/> and we found it was zero <pause dur="0.3"/> so if we have an algebraic integer <pause dur="0.4"/> it's either one <pause dur="0.4"/> or zero <pause dur="0.2"/> in absolute value <pause dur="3.5"/> so that's a <pause dur="1.5"/> a bit of preparation <pause dur="0.3"/> and now <pause dur="0.9"/> the next result is <pause dur="1.9"/> right at the core <pause dur="0.8"/> of <pause dur="0.9"/> the proof of Burnside's theorem <pause dur="0.3"/> it's the really clever bit <pause dur="0.4"/> so this is the bit you have to watch carefully <pause dur="0.4"/> and make sure that i've got nothing <pause dur="0.3"/> up my sleeve <pause dur="1.1"/><vocal desc="clears throat" iterated="n"/><pause dur="12.4"/> so we'll call this one sixteen-point-two <pause dur="1.1"/><kinesic desc="writes on board" iterated="y" dur="15"/> and it's sufficiently important to be called a theorem i think <pause dur="1.0"/> and it was due to Burnside <pause dur="4.9"/> # around <pause dur="0.6"/>

nineteen-hundred <pause dur="1.7"/> long time ago <pause dur="0.5"/> they were clever even then these guys <pause dur="0.5"/> these group theorists <pause dur="0.9"/> # <pause dur="1.4"/> now what does it say <pause dur="0.6"/> # <pause dur="0.3"/> suppose G <pause dur="2.8"/><kinesic desc="writes on board" iterated="y" dur="3"/> that's a finite group G <pause dur="1.6"/> has a conjugacy class <pause dur="0.8"/> with <pause dur="0.5"/> P-to-the-R <pause dur="0.6"/> elements <pause dur="9.2"/><kinesic desc="writes on board" iterated="y" dur="9"/> so that stands for the conjugacy class <pause dur="1.1"/><vocal desc="clears throat" iterated="n"/><pause dur="3.0"/><kinesic desc="writes on board" iterated="y" dur="17"/> so the number of conjugates is P-to-the-R <pause dur="0.4"/> for some prime P <pause dur="3.7"/> and <pause dur="1.7"/> some natural number R <pause dur="0.4"/><vocal desc="clears throat" iterated="n"/><pause dur="0.5"/> greater than or equal to one <pause dur="5.6"/> notice straight away <pause dur="0.6"/><kinesic desc="indicates point on board" iterated="n"/> this hypothesis tells me that the group is not abelian <pause dur="1.3"/> because in a <trunc>c</trunc> <pause dur="0.8"/> an abelian group <pause dur="0.4"/> all the conjugacy class <trunc>c</trunc> classes contain one element <pause dur="27.7"/><kinesic desc="writes on board" iterated="y" dur="7"/> now we need a conclusion we've got a hypothesis <pause dur="0.4"/> the conclusion is <pause dur="0.4"/> that G <pause dur="5.4"/><kinesic desc="writes on board" iterated="y" dur="8"/> is not a simple group <pause dur="3.5"/> and the definition of a simple group is one with exactly two normal subgroups <pause dur="0.9"/><vocal desc="clears throat" iterated="n"/><pause dur="8.6"/><kinesic desc="writes on board" iterated="y" dur="17"/>

it has a proper <pause dur="3.9"/> non-trivial <pause dur="3.8"/> if i can just squeeze it in here <pause dur="0.4"/> so that you can read it <pause dur="3.6"/><kinesic desc="writes on board" iterated="y" dur="6"/> normal <pause dur="1.9"/> subgroup <pause dur="4.2"/> i've got to find some <pause dur="0.7"/> proper normal subgroup which is bigger than one <pause dur="2.5"/><vocal desc="clears throat" iterated="n"/><pause dur="2.5"/> i'm going to use <pause dur="1.2"/> just sixteen-point-one <pause dur="4.7"/><vocal desc="clears throat" iterated="n"/><pause dur="1.4"/> and the first thing i do <pause dur="2.0"/> is to <pause dur="0.8"/><vocal desc="sniff" iterated="n"/><pause dur="5.2"/> apply so this is the <kinesic desc="writes on board" iterated="y" dur="3"/> proof coming up <pause dur="1.1"/> very delicate proof <pause dur="2.0"/> i apply the orthogonality relationship <pause dur="1.5"/><kinesic desc="writes on board" iterated="y" dur="2"/> apply <pause dur="1.0"/> it will be <pause dur="0.3"/> a column <pause dur="0.6"/> orthogonality <kinesic desc="writes on board" iterated="y" dur="12"/> relation <pause dur="7.3"/> to column <pause dur="2.9"/> one that's the <pause dur="0.7"/> one that has the degrees of the character in it corresponding to the <pause dur="0.6"/> conjugacy class of the identity element <pause dur="0.3"/> and <pause dur="3.0"/><kinesic desc="writes on board" iterated="y" dur="6"/>

the column <pause dur="2.8"/> of G <pause dur="0.4"/> G is the representative of the <pause dur="0.5"/> class <pause dur="1.2"/> with <pause dur="2.2"/> P-to-the-R <pause dur="6.0"/><kinesic desc="writes on board" iterated="y" dur="6"/> and what do i get well nought <pause dur="0.2"/> we know the formula by now <pause dur="1.0"/><kinesic desc="writes on board" iterated="y" dur="1"/> the inner product <pause dur="0.3"/> is # zero <pause dur="0.7"/> and # <pause dur="1.1"/> i have the <pause dur="1.1"/><kinesic desc="writes on board" iterated="y" dur="2"/> sum <pause dur="0.4"/> over the irreducible characters <pause dur="0.3"/> running down the column <pause dur="0.9"/><kinesic desc="writes on board" iterated="y" dur="3"/> of the character table <pause dur="2.4"/> of # <pause dur="0.4"/> well <pause dur="1.4"/><kinesic desc="writes on board" iterated="y" dur="1"/> i can <pause dur="0.3"/> it's it's symmetric so i can write # chi-<pause dur="0.4"/><kinesic desc="writes on board" iterated="y" dur="2"/>G <pause dur="0.9"/> chi-one-bar but <pause dur="0.4"/> chi-one is a natural number so chi-one-bar is chi-one <pause dur="0.4"/> and i i write this <pause dur="0.4"/> as # <pause dur="0.5"/><kinesic desc="writes on board" iterated="y" dur="1"/> one corresponding to the trivial character <pause dur="0.7"/><kinesic desc="writes on board" iterated="y" dur="4"/> plus the <pause dur="0.2"/> sum <pause dur="0.6"/> from I-equals-two up to the class number K <pause dur="1.0"/> of

# <pause dur="1.6"/> <trunc>g</trunc> <pause dur="1.3"/><kinesic desc="writes on board" iterated="y" dur="7"/> chi-<pause dur="1.8"/>I-G <pause dur="1.1"/> chi-<pause dur="0.7"/>I-<pause dur="0.2"/>one <pause dur="0.8"/> so <pause dur="0.5"/> in this sum <pause dur="1.6"/> i've taken away the <pause dur="1.1"/> trivial character chi-one <pause dur="0.4"/> and i've listed all the other irreducible characters chi-two up to chi-K <pause dur="0.3"/> and we know the number in all is <pause dur="0.4"/> little-K the class number of the group <pause dur="0.8"/><vocal desc="sniff" iterated="n"/><pause dur="2.0"/><vocal desc="clears throat" iterated="n"/><pause dur="1.9"/> so <pause dur="0.2"/> this is all <pause dur="3.8"/> it looks tricky but it's it's it's really rather nice so <pause dur="0.3"/> i <trunc>d</trunc> i divide by P because # <pause dur="0.6"/> well you'll see why i divide by P in a minute <pause dur="0.5"/> take this over to the other side divide by P <pause dur="2.5"/><kinesic desc="writes on board" iterated="y" dur="3"/> and # <pause dur="1.4"/> what do i get <pause dur="0.3"/><vocal desc="clears throat" iterated="n"/><pause dur="14.7"/><kinesic desc="writes on board" iterated="y" dur="11"/> i write it like that <pause dur="1.2"/> the sum <pause dur="0.2"/> from <pause dur="1.4"/><kinesic desc="writes on board" iterated="y" dur="3"/> K-equals-two <pause dur="0.9"/> to K <pause dur="0.6"/> remember these are all non-trivial <pause dur="0.6"/> irreducible characters <pause dur="0.9"/> of this product is <pause dur="0.9"/> minus-one-upon-P P's a prime <pause dur="0.6"/> so the right-hand side is therefore <pause dur="0.4"/> not an algebraic <pause dur="0.6"/> integer <pause dur="4.6"/> because the only <pause dur="0.5"/> rationals that are algebraic integers are the <pause dur="1.1"/> so-called rational

integers the ordinary integers themselves </u><pause dur="0.8"/> <u who="sm0943" trans="pause"> <gap reason="inaudible" extent="1 sec"/></u><pause dur="0.3"/> <u who="nm0941" trans="pause"> have i got </u><u who="sm0943" trans="latching"> <gap reason="inaudible" extent="1 sec"/> </u><pause dur="0.9"/> <u who="nm0941" trans="pause"> # <pause dur="0.5"/> ah thank you <kinesic desc="writes on board" iterated="y" dur="2"/><pause dur="19.8"/><kinesic desc="waves to member of audience" iterated="n"/> <vocal desc="laughter" iterated="y" n="sl" dur="2"/><pause dur="0.5"/> that really threw you that one i can see <pause dur="0.7"/><vocal desc="laughter" iterated="y" n="ss" dur="1"/> so the right-hand side <pause dur="3.5"/><kinesic desc="writes on board" iterated="y" dur="8"/> not <pause dur="0.8"/> an algebraic integer <pause dur="0.6"/> surprise surprise <pause dur="1.4"/> therefore the left-hand side <pause dur="0.7"/> is not an algebraic integer <pause dur="1.7"/> now you have to remember a fact i proved <pause dur="0.2"/> no it's a fact i didn't prove in fact <pause dur="0.9"/> namely <pause dur="0.4"/> that the <pause dur="0.2"/> algebraic integers form a ring and therefore in particular <pause dur="0.4"/> # sums of products of

algebraic integers are algebraic integers <pause dur="0.3"/> and on the left-hand side here <pause dur="0.5"/> i have <pause dur="0.5"/> # a sum of products <pause dur="0.8"/> so <pause dur="0.3"/><vocal desc="clears throat" iterated="n"/><pause dur="0.2"/><kinesic desc="indicates point on board" iterated="n"/> this is an algebraic integer here <pause dur="3.6"/><kinesic desc="writes on board" iterated="y" dur="3"/> and <pause dur="0.3"/> if the sum of these products is to be an <trunc>al</trunc> is is not to be an algebraic integer <pause dur="0.3"/> at least <kinesic desc="indicates point on board" iterated="n"/> one of these things <pause dur="1.0"/> <kinesic desc="writes on board" iterated="y" dur="1"/> must not be <pause dur="0.9"/> an algebraic integer <pause dur="8.0"/><kinesic desc="writes on board" iterated="y" dur="10"/> there exists some <pause dur="0.3"/> I <pause dur="1.0"/> between two <pause dur="1.3"/> and K <pause dur="1.9"/> <trunc>a</trunc> and of course <pause dur="0.9"/> # <pause dur="8.4"/> yeah <pause dur="5.8"/> it's no good having <pause dur="0.5"/><kinesic desc="indicates point on board" iterated="n"/> this zero because that won't kill things off <pause dur="0.3"/> and <pause dur="1.1"/><kinesic desc="writes on board" iterated="y" dur="20"/> # <pause dur="1.3"/> chi-I-G <pause dur="0.4"/> not equal to zero <pause dur="2.4"/> such that <pause dur="2.5"/> chi-I-one divided by P <pause dur="0.5"/> is again <pause dur="0.2"/> not <pause dur="0.9"/> an algebraic <pause dur="1.1"/> integer <pause dur="2.6"/> if all of <kinesic desc="indicates point on board" iterated="n"/> these for non-zero values of that were algebraic integers then the left-hand side would be an algebraic integer <pause dur="0.4"/> that would be a contradiction <pause dur="3.9"/> in other words <pause dur="7.1"/><kinesic desc="writes on board" iterated="y" dur="4"/> P <pause dur="0.6"/> does not

divide <pause dur="2.9"/><kinesic desc="writes on board" iterated="y" dur="3"/> the degree <pause dur="1.1"/> of this character chi-I <pause dur="2.8"/> if it did then this would be an integer <pause dur="0.3"/> therefore an algebraic integer <pause dur="4.0"/> at this point <pause dur="0.2"/> you <pause dur="0.2"/> cast your mind back to <pause dur="0.6"/> first year foundations did any of you here do it with me <pause dur="1.7"/><kinesic desc="put hands up" iterated="n" n="ss"/> oh faithful <pause dur="0.5"/> that you are <pause dur="0.8"/><vocal desc="laughter" iterated="y" dur="1"/><pause dur="0.3"/> still hanging in there <pause dur="0.7"/> we did something called # <pause dur="0.8"/> # a consequence of the Euclidean algorithm then <pause dur="1.5"/> and # <pause dur="0.8"/><vocal desc="clears throat" iterated="n"/><pause dur="2.9"/> you see what is <pause dur="2.3"/> yeah <pause dur="1.1"/> P <pause dur="1.4"/> is the power <pause dur="0.4"/><event desc="drops board rubber" iterated="n"/> whoops <pause dur="0.6"/> P is the power of the prime <pause dur="0.4"/> for which <pause dur="1.4"/> G belongs to a conjugacy class <pause dur="0.5"/> with P-to-the-R elements <pause dur="0.4"/> # <pause dur="0.2"/> hence <pause dur="1.9"/><kinesic desc="writes on board" iterated="y" dur="19"/> P <pause dur="0.2"/> and <pause dur="1.2"/> the number of elements in this conjugacy class <pause dur="1.0"/> sorry <pause dur="0.7"/> i should say there chi-of-one <pause dur="0.8"/> chi-I-of-one <pause dur="3.4"/> and this which is <pause dur="1.4"/> P-to-the-R but <pause dur="0.2"/> if you remember the other formula for the number of elements in a conjugacy class <pause dur="0.5"/> it's

precisely <pause dur="0.4"/><kinesic desc="writes on board" iterated="y" dur="10"/> the order of G <pause dur="0.5"/> divided by <pause dur="1.3"/> the order of the centralizer that was a consequence of the orbit-stabilizer theorem <pause dur="1.4"/> so these two numbers are coprime <pause dur="7.6"/> by <pause dur="1.7"/> the Euclidean algorithm <pause dur="1.4"/> we know there exist <pause dur="2.1"/><kinesic desc="writes on board" iterated="y" dur="3"/> integers A and B <pause dur="0.9"/> such that <pause dur="2.5"/><kinesic desc="writes on board" iterated="y" dur="1"/> the highest common factor of <kinesic desc="indicates point on board" iterated="n"/> these two which is one <pause dur="3.2"/><kinesic desc="writes on board" iterated="y" dur="1"/> is <pause dur="0.8"/> A times <pause dur="1.0"/><kinesic desc="writes on board" iterated="y" dur="9"/> the order of G divided by <pause dur="1.0"/> the order of the centralizer of little-G <pause dur="1.0"/> # <pause dur="0.8"/> plus <pause dur="0.7"/> B times <pause dur="1.1"/> what do we have chi of <pause dur="0.3"/><kinesic desc="writes on board" iterated="y" dur="6"/> chi-I-of-one a degree <pause dur="0.8"/> this is equal to the highest common factor <pause dur="1.0"/> two coprime integers you can always find A and B so that equation holds <pause dur="3.1"/> okay <pause dur="5.5"/> now we slightly manipulate this equation <pause dur="1.0"/> # <pause dur="0.2"/><kinesic desc="writes on board" iterated="y" dur="4"/> multiply by <pause dur="2.9"/> chi </u><pause dur="1.4"/> <u who="sm0944" trans="pause"> excuse me on the subscript for the centralizer </u><pause dur="0.4"/> <u who="nm0941" trans="pause"> oh yeah have i got it wrong </u><u who="sm0944" trans="latching"> you've got I and G which one is it </u><pause dur="0.4"/> <u who="nm0941" trans="pause"> # <pause dur="1.1"/> oh it's <trunc>j</trunc> it's big-G it's a centralizer in the group yeah thank you <pause dur="8.7"/><kinesic desc="writes on board" iterated="y" dur="1"/> multiply <pause dur="1.7"/>

times <kinesic desc="writes on board" iterated="y" dur="30"/> by <pause dur="1.1"/> well what do we want # <pause dur="1.0"/> chi-I-of-G <pause dur="0.8"/> over chi-<pause dur="0.6"/>I-of-one <pause dur="2.1"/> to get <pause dur="1.0"/> now <pause dur="1.0"/> if things work out well <pause dur="1.2"/> the left-hand side is going to be <pause dur="2.6"/> in the form <pause dur="1.1"/> of something which we proved last time <pause dur="2.9"/> was an algebraic integer <pause dur="2.8"/> and then what have we got there <pause dur="0.2"/> we've got <pause dur="0.2"/> that's A times that <pause dur="1.3"/><kinesic desc="writes on board" iterated="y" dur="10"/> and B times <pause dur="1.5"/> # <pause dur="1.4"/> chi-I-of-G <pause dur="0.5"/> upon <pause dur="0.6"/> chi-I-of-one <pause dur="1.0"/> this is equal to <pause dur="1.4"/> # <pause dur="1.5"/> have i got that right no <pause dur="0.8"/> i've forgotten to cancel out <pause dur="1.0"/> # <pause dur="3.2"/> so that's equal to <kinesic desc="writes on board" iterated="y" dur="2"/> one times <pause dur="0.7"/> chi-I-of-<pause dur="1.7"/>G <pause dur="1.1"/><kinesic desc="writes on board" iterated="y" dur="3"/> over chi-I-of-one <pause dur="0.3"/> now i've put it in that form <pause dur="2.5"/> for the following reason look at the left-hand side <pause dur="0.8"/> # <pause dur="0.3"/> A <pause dur="0.6"/> is an integer <pause dur="2.3"/><kinesic desc="writes on board" iterated="y" dur="2"/> that particular expression we showed last time <pause dur="1.5"/> in <pause dur="1.4"/> fifteen-point-one <pause dur="1.5"/><vocal desc="clears throat" iterated="n"/><pause dur="1.6"/><kinesic desc="writes on board" iterated="y" dur="4"/> is an algebraic <pause dur="0.5"/> integer <pause dur="5.5"/> # <pause dur="0.2"/> we know the character values

are algebraic integers <pause dur="1.1"/><kinesic desc="writes on board" iterated="y" dur="10"/> hence <pause dur="1.5"/> the left-hand side <pause dur="0.9"/> is <pause dur="1.0"/> another algebraic integer <pause dur="3.0"/> being a sum of products of algebraic integers <pause dur="16.6"/> now look at the right-hand side <pause dur="1.2"/> and look at sixteen-point-one in your notes i've rubbed it out now <pause dur="10.9"/> by the first result we proved <pause dur="0.5"/> today <pause dur="5.9"/><kinesic desc="writes on board" iterated="y" dur="2"/><vocal desc="sniff" iterated="n"/><pause dur="1.1"/><kinesic desc="indicates point on board" iterated="n"/> this is an algebraic integer <pause dur="1.2"/> and the result said <pause dur="0.4"/><kinesic desc="indicates point on board" iterated="n"/> this is either zero or one <pause dur="0.6"/> but the bottom <kinesic desc="indicates point on board" iterated="n"/> here's a natural number <pause dur="0.5"/> the top <kinesic desc="indicates point on board" iterated="n"/> here is non-zero by choice <pause dur="1.1"/> so it must be one <pause dur="12.4"/><kinesic desc="writes on board" iterated="y" dur="6"/> so we manage to pick out <pause dur="0.2"/> an irreducible character <pause dur="0.7"/><vocal desc="sniff" iterated="n"/><pause dur="2.6"/> whose value <pause dur="1.3"/> on a non-trivial element <pause dur="1.7"/> coincides with <pause dur="0.3"/> the value <pause dur="0.8"/> on the identity which is the degree <pause dur="2.2"/> and we've already proved something about that <pause dur="1.3"/><vocal desc="clears throat" iterated="n"/><pause dur="5.5"/> # <pause dur="1.0"/><vocal desc="sniff" iterated="n"/><pause dur="4.3"/><kinesic desc="writes on board" iterated="y" dur="3"/> # <pause dur="0.2"/> well <trunc>l</trunc> <trunc>l</trunc> before i go to case one let <pause dur="1.8"/><kinesic desc="writes on board" iterated="y" dur="14"/> rho <pause dur="0.5"/> be <pause dur="1.5"/> the <pause dur="0.8"/> representation <pause dur="1.0"/> it's going to be an irreducible representation <pause dur="0.5"/> affording <pause dur="1.1"/>

this irreducible character chi-I <pause dur="0.6"/> so chi-I is obtained by taking the trace <pause dur="0.4"/> of the matrices <pause dur="2.6"/> which are <pause dur="0.5"/> images under rho <pause dur="1.9"/><vocal desc="sniff" iterated="n"/><pause dur="1.5"/> so <kinesic desc="writes on board" iterated="y" dur="3"/> case one <pause dur="3.6"/> # <pause dur="0.9"/><kinesic desc="writes on board" iterated="y" dur="6"/> rho is not faithful <pause dur="4.6"/> that means the kernel is non-trivial <pause dur="2.8"/> and the kernel can't be the whole of G because <pause dur="0.3"/> chi-I is not equal to chi-one <pause dur="19.9"/><kinesic desc="writes on board" iterated="y" dur="19"/><vocal desc="sniff" iterated="n"/><pause dur="4.0"/> the only time the kernel is the whole group is when we have the trivial representation <pause dur="0.7"/> and the trivial character <pause dur="1.1"/> and so <pause dur="0.3"/> we're done in this case <pause dur="0.8"/><kinesic desc="writes on board" iterated="y" dur="2"/> we have <pause dur="1.6"/> the kernel <pause dur="0.2"/> as a non-trivial <pause dur="1.3"/> proper <pause dur="0.2"/> normal <pause dur="0.2"/> subgroup <pause dur="3.7"/> are you following <pause dur="1.6"/> everyone following </u><pause dur="0.9"/> <u who="sm0945" trans="pause"> <gap reason="inaudible" extent="1 sec"/> </u><pause dur="0.5"/> <u who="nm0941" trans="pause"> yeah <pause dur="0.2"/> question good </u><pause dur="0.2"/> <u who="sm0945" trans="pause"> <gap reason="inaudible" extent="1 sec"/> <vocal desc="laughter" iterated="y" dur="1"/><pause dur="0.3"/> <gap reason="inaudible" extent="4 secs"/></u><pause dur="0.4"/> <u who="nm0941" trans="pause"> on <kinesic desc="indicates board" iterated="n"/> this board </u><u who="sm0945" trans="latching"> <trunc>bo</trunc> bottom right <pause dur="0.9"/> bottom right </u><pause dur="0.3"/> <u who="nm0941" trans="pause"> in other words sorry </u><u who="sm0945" trans="overlap"> <gap reason="inaudible" extent="1 sec"/> <vocal desc="laughter" iterated="y" n="ss" dur="1"/> </u><pause dur="0.8"/> <u who="nm0941" trans="pause"> P does not divide </u><pause dur="0.3"/> <u who="sm0945" trans="pause">

bottom right board <kinesic desc="indicates board" iterated="n"/> </u><pause dur="0.2"/> <u who="nm0941" trans="pause"> <trunc>bo</trunc> oh bottom right board okay okay <vocal desc="laughter" iterated="y" n="sl" dur="1"/> </u><pause dur="1.1"/> <u who="sm0945" trans="pause"> you have <gap reason="inaudible" extent="2 secs"/> in the brackets yeah </u><pause dur="0.4"/> <u who="nm0941" trans="pause"> # sorry it's yeah it's that thing <kinesic desc="writes on board" iterated="y" dur="3"/> thanks yeah </u><u who="sm0945" trans="latching"><gap reason="inaudible" extent="1 sec"/> </u><pause dur="0.5"/> <u who="nm0941" trans="pause"> sorry <pause dur="1.1"/> that's exactly what fifteen-point-one <pause dur="0.2"/> says okay that that thing <pause dur="0.3"/> <kinesic desc="indicates point on board" iterated="n"/> this is an integer <kinesic desc="indicates point on board" iterated="n"/> that's an algebraic integer <kinesic desc="indicates point on board" iterated="n"/> that's an integer <pause dur="0.3"/> <kinesic desc="indicates point on board" iterated="n"/> that's an algebraic integer there for the whole of <kinesic desc="indicates point on board" iterated="n"/> this <pause dur="0.7"/><vocal desc="sniff" iterated="n"/><pause dur="0.7"/><vocal desc="sniff" iterated="n"/><pause dur="1.7"/> so what's the other case <pause dur="1.0"/> # <pause dur="3.4"/> oh i'm proving <kinesic desc="indicates point on board" iterated="n"/> that so i won't rub it out <pause dur="10.0"/> well the other case is where <pause dur="0.4"/><vocal desc="clears throat" iterated="n"/><pause dur="2.6"/> the the representation <kinesic desc="writes on board" iterated="y" dur="6"/> is faithful <pause dur="2.4"/><vocal desc="clears throat" iterated="n"/><pause dur="2.2"/> the kernel's one <pause dur="1.5"/> and so in this case <pause dur="0.7"/> the image of rho <pause dur="0.2"/> is isomorphic to G itself <pause dur="6.4"/><kinesic desc="writes on board" iterated="y" dur="4"/> well <pause dur="0.3"/> if you go back to <pause dur="0.9"/> eight-point-four <pause dur="0.6"/> it's a long

time ago now but you may <pause dur="0.6"/> have a distant recollection <pause dur="1.1"/> </u><pause dur="1.2"/> <u who="sm0946" trans="pause"> <gap reason="inaudible" extent="1 sec"/> </u><pause dur="0.3"/> <u who="nm0941" trans="pause"> mm-hmm </u><pause dur="0.6"/> <u who="sm0946" trans="pause"> # you put # chi-I not equal to <pause dur="0.2"/> chi-one </u><pause dur="0.4"/> <u who="nm0941" trans="pause"> chi-I equals # not sorry not equal to chi-one yeah <pause dur="0.8"/><kinesic desc="writes on board" iterated="y" dur="1"/> you're <trunc>r</trunc> really sharp today thank you <pause dur="0.2"/> i'm not so sharp but you are <pause dur="0.3"/> # <pause dur="1.4"/> sorry about that <pause dur="2.5"/> # <pause dur="1.1"/> if it's faithful well <pause dur="1.0"/> by <pause dur="1.9"/><kinesic desc="writes on board" iterated="y" dur="2"/> if you look back eight-point-four-<pause dur="0.3"/>A <pause dur="1.8"/> # <pause dur="1.7"/> the <pause dur="5.2"/> the statement there was that # <pause dur="0.7"/> rho-of-G <pause dur="3.9"/><kinesic desc="writes on board" iterated="y" dur="2"/> <vocal desc="clears throat" iterated="n"/><pause dur="0.5"/> <trunc>i</trunc> <trunc>i</trunc> <trunc>i</trunc> <pause dur="0.6"/> it's <kinesic desc="indicates point on board" iterated="n"/> this <kinesic desc="writes on board" iterated="y" dur="2"/> condition star here <pause dur="1.0"/> using star <pause dur="3.9"/><kinesic desc="writes on board" iterated="y" dur="3"/> that was the hypothesis of eight-point-four-A it said in this condition <pause dur="0.6"/> then <pause dur="0.5"/> the representation affording <pause dur="0.8"/> chi-I <pause dur="1.0"/> represents G as a scalar matrix <pause dur="11.9"/><kinesic desc="writes on board" iterated="y" dur="7"/> so <pause dur="2.8"/><kinesic desc="writes on board" iterated="y" dur="4"/> hence <pause dur="1.0"/> rho-of-G <pause dur="0.9"/> because scalar matrices commute with all matrices <pause dur="0.9"/> that are D by D <pause dur="1.1"/> # <pause dur="0.6"/><kinesic desc="writes on board" iterated="y" dur="12"/> that's

in the centre <pause dur="2.9"/> of <pause dur="1.0"/> rho-of-G <pause dur="0.9"/> and it's not hard to see that that that's the same <pause dur="0.7"/> since rho is an isomorphism <pause dur="1.1"/> it's the image of <pause dur="0.3"/> the centre of G <pause dur="1.7"/> so apply <pause dur="0.4"/> the inverse map for rho <pause dur="0.7"/> since it's bijective <pause dur="6.6"/><vocal desc="sniff" iterated="n"/><pause dur="18.5"/><kinesic desc="writes on board" iterated="y" dur="18"/> and that says that the group has non-trivial centre <pause dur="2.7"/><vocal desc="sniff" iterated="n"/><pause dur="0.3"/> now we saw at the very beginning that the hypothesis <pause dur="0.5"/> implied that G was not abelian <pause dur="17.7"/><kinesic desc="writes on board" iterated="y" dur="12"/> # <pause dur="0.2"/> Z-G <pause dur="1.5"/> is a proper Z-G is only G if the group's abelian <pause dur="0.5"/> is a proper <pause dur="2.6"/><kinesic desc="writes on board" iterated="y" dur="12"/> non-trivial <pause dur="2.4"/> normal subgroup <pause dur="6.9"/> okay well <pause dur="0.7"/> take a <pause dur="0.2"/> deep breath that was quite a bit of hard work you've done very well to <pause dur="0.7"/> keep <pause dur="0.4"/> your attention going i've got <pause dur="0.4"/> i just want to give you a short announcement <pause dur="0.7"/> about # <pause dur="1.8"/> ooh what bits of paper you're likely to get <pause dur="0.9"/> today you'll get something called <pause dur="0.5"/> Aims Objectives <pause dur="0.4"/> and Syllabus <pause dur="0.6"/> you may find it helpful to

see what i was trying to do and what i've covered <pause dur="0.7"/> it's certainly <pause dur="0.6"/> # de rigueur to distribute one of these before the end of the course <pause dur="0.6"/> # on Friday <pause dur="0.6"/> i hope <pause dur="0.8"/> # <pause dur="0.6"/> very much earnestly to have had time to write you <pause dur="0.5"/> at least one <pause dur="0.2"/> mock exam paper so <pause dur="0.4"/> if you can't get along on Friday because Christmas calls <pause dur="0.5"/> then <pause dur="0.3"/> as soon as i've done it i'll stick <pause dur="0.3"/> some copies in the filing cabinet in the atrium yeah </u><pause dur="0.3"/> <u who="sm0947" trans="pause"> # with <pause dur="0.2"/> respect to <gap reason="inaudible" extent="1 sec"/> that you <trunc>n</trunc> you not recommend using <gap reason="inaudible" extent="1 sec"/> </u><u who="nm0941" trans="overlap"> # i've had a look at last year's that <gap reason="name" extent="2 words"/> set <pause dur="0.2"/> # <pause dur="0.6"/> yeah there are some bits and pieces there i mean what i would advise you to do have a look at it <pause dur="0.5"/> see if it figures <pause dur="0.7"/> he's more or less covered <pause dur="0.6"/> about half of what i've covered but from <pause dur="0.4"/> a different emphasis i've put more emphasis on character tables <pause dur="0.4"/> so <pause dur="0.3"/> you can make sense of some of the stuff there <pause dur="0.4"/> do what you can <pause dur="0.3"/> and i'll <pause dur="0.5"/> produce <pause dur="0.2"/> # <pause dur="0.7"/> my own mock exam and i will also <pause dur="0.6"/> put up the solutions next term <pause dur="0.2"/> on <unclear>maths stuff</unclear> so if you

<trunc>h</trunc> are struggling with those questions <pause dur="0.4"/> by a couple of weeks into the term you should find <pause dur="0.5"/> the solutions on <unclear>maths stuff</unclear> <pause dur="0.3"/> so i i do i have got a few more things to do but <pause dur="0.3"/> i'll just send these round <pause dur="0.5"/> while <pause dur="0.2"/> # <pause dur="0.7"/> while i clean the boards <pause dur="1.3"/> would <pause dur="1.1"/> it's a bit <pause dur="0.2"/> would you mind just <pause dur="0.2"/> going up this <pause dur="0.2"/> this row <event desc="passes out handouts" iterated="n"/></u><pause dur="4.1"/> <u who="sm0948" trans="pause"> excuse me </u><pause dur="0.2"/> <u who="nm0941" trans="pause"> mm </u><pause dur="0.4"/> <u who="sm0948" trans="pause"> do you know when the test # when the <gap reason="inaudible" extent="1 sec"/></u><u who="nm0941" trans="overlap"> exam it's in April i think </u><u who="sm0948" trans="overlap"> no no no </u><gap reason="break in recording" extent="uncertain"/> <u who="nm0941" trans="pause"> okay in the last few minutes i want to just prove something which i think you may have <pause dur="0.5"/> seen already but i'll just <pause dur="0.3"/> use it as revision <pause dur="8.0"/><kinesic desc="writes on board" iterated="y" dur="9"/> if <pause dur="0.3"/> you have a finite group <pause dur="2.4"/> # <pause dur="1.9"/> whose order is the power of some prime <pause dur="4.4"/><kinesic desc="writes on board" iterated="y" dur="4"/> then the centre <pause dur="0.5"/> if it's if it's a <pause dur="0.5"/> a non-trivial group <kinesic desc="writes on board" iterated="y" dur="5"/> then the

centre <pause dur="2.4"/> of G <pause dur="0.2"/> is also bigger than one <pause dur="7.2"/><kinesic desc="writes on board" iterated="y" dur="2"/> well G is a union of conjugacy classes <pause dur="14.2"/><kinesic desc="writes on board" iterated="y" dur="13"/> so when you add them all up you get <pause dur="0.8"/> the number of elements in the group is P-to-the-M <pause dur="1.4"/><kinesic desc="writes on board" iterated="y" dur="12"/> this is the order of G <pause dur="0.6"/> and then this is <pause dur="0.6"/> one-to-the-G <pause dur="1.7"/> G-two-to-the-G <pause dur="5.6"/> where <pause dur="0.9"/> one G-two <pause dur="0.7"/> G-three and so on are just representatives <pause dur="0.5"/> of the conjugacy classes <pause dur="1.7"/> now <pause dur="1.9"/> the number of conjugates are <pause dur="1.0"/><kinesic desc="writes on board" iterated="y" dur="1"/> the identity is one <pause dur="1.2"/> and <pause dur="1.4"/><kinesic desc="writes on board" iterated="y" dur="2"/> the other conjugates are <kinesic desc="writes on board" iterated="y" dur="4"/> P-to-the-M-two <pause dur="0.7"/> plus <pause dur="0.5"/> P-to-the-M-three <pause dur="1.0"/> plus <pause dur="0.2"/> so on because <pause dur="0.6"/> the number of <pause dur="0.2"/> elements in a conjugacy class is the index of the centralizer of an element <pause dur="0.4"/> so it divides the order of the group <pause dur="6.2"/> so the left-hand side <pause dur="0.8"/> is congruent to zero-<pause dur="0.5"/>modulo-P <pause dur="7.3"/><kinesic desc="writes on board" iterated="y" dur="12"/> if <pause dur="1.8"/> M-two <pause dur="0.8"/> M-three and so on <pause dur="0.8"/> all <pause dur="0.2"/> bigger than one <pause dur="1.0"/> then <pause dur="1.3"/> the right-hand side <pause dur="1.0"/> is congruent to one-mod-P <pause dur="8.5"/><kinesic desc="writes on board" iterated="y" dur="5"/> but that's <pause dur="1.7"/> nonsense because the left-hand side <pause dur="3.9"/><kinesic desc="writes on board" iterated="y" dur="1"/>

is congruent to zero-mod-P <pause dur="1.3"/> and you can't have that when P is the prime <pause dur="1.0"/> and so that's a contradiction there's my <pause dur="1.0"/><kinesic desc="indicates point on board" iterated="n"/> favourite little sign for a contradiction <pause dur="2.1"/> so <pause dur="1.0"/><kinesic desc="writes on board" iterated="y" dur="5"/> some <pause dur="1.3"/> M-I <pause dur="0.4"/><vocal desc="clears throat" iterated="n"/><pause dur="1.9"/> i should say if they're greater than or equal to <kinesic desc="writes on board" iterated="y" dur="1"/> one i guess <pause dur="0.2"/> yep <pause dur="0.5"/><vocal desc="sniff" iterated="n"/><pause dur="0.8"/> so they can't all be greater than or equal to one so some M-I is zero <pause dur="0.7"/><kinesic desc="writes on board" iterated="y" dur="6"/> then <pause dur="0.8"/> <trunc>m</trunc> <pause dur="1.3"/> mod-<pause dur="2.6"/>G-I-to-the-G <pause dur="0.8"/> is one <pause dur="0.9"/><kinesic desc="writes on board" iterated="y" dur="2"/> P-to-the-zero which is one <pause dur="3.2"/> hence <kinesic desc="writes on board" iterated="y" dur="7"/></u><u who="sm0949" trans="latching"> you've got contradiction <gap reason="inaudible" extent="1 sec"/></u><pause dur="3.0"/> <u who="sm0950" trans="pause"> that <pause dur="0.2"/> little <pause dur="0.2"/> lightning bolt's the </u><u who="nm0941" trans="overlap"> that is a contradiction right <pause dur="0.4"/> yeah <pause dur="7.0"/> <kinesic desc="writes on board" iterated="y" dur="5"/> i might just as well have written it out </u><pause dur="1.7"/> <u who="sm0949" trans="pause"> <gap reason="inaudible" extent="2 secs"/></u><u who="nm0941" trans="latching"> <vocal desc="laugh" iterated="n"/> it is a contradiction one side's

congruent to nought-mod-P the other side's congruent to one-mod-P you can't get more <pause dur="0.4"/> contradictory than that can you <vocal desc="laughter" iterated="y" n="ss" dur="1"/> i mean <pause dur="0.4"/> life would collapse <pause dur="0.9"/> # <pause dur="1.0"/> so <pause dur="0.5"/> what does it mean to say there's one conjugate <pause dur="0.9"/> of G-I it means whenever i conjugate by anything in <pause dur="0.5"/> G it stays the same <pause dur="1.3"/><kinesic desc="writes on board" iterated="y" dur="7"/> that means it commutes with everything <pause dur="2.1"/> so it's in the centre of G <pause dur="1.9"/> it was not the identity because we separated the identity off <pause dur="0.8"/> and that <pause dur="0.5"/> gives me the result <pause dur="4.1"/><kinesic desc="writes on board" iterated="y" dur="2"/> good well we're in <trunc>g</trunc> we're we're in pretty good shape for our climax on Friday <pause dur="0.5"/> so i look forward to seeing you then

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