nm0941: the object of the last two lectures of this course today and Friday are 
to send you home in a good mood er on Friday we shall reach a climax with a 
proof of a theorem i've already trailered namely Burnside's so-called P-to-the-
A Q-to-the-B theorem and today we shall indulge in a little foreplay right 
we're going to u-, exploit some of the facts we have stated about algebraic 
numbers if you have a character of a group G and you have an element of the 
group G then a character value on G divided by the degree of that character is 
actually an algebraic number and er it satisfies that's the easy bit the 
important bit is that er if it's an algebraic integer then the absolute value 
of this 
quotient is nought or one well we know er let's call that thing alpha to save 
having to keep writing it out that's certainly a complex number it's clearly an 
algebraic integer because we know that the character values or sums of roots of 
unity and i'm just dividing by an integer er so this is equal to yes then er 
chi-of-G is a sum of roots of unity er therefore if i look at the absolute 
value of alpha this is er the absolute value of this divided by D by the 
triangle of er complex numbers that's less than or equal to the absolute value 
of the roots of unity the absolute value of a root of unity they lie on the 
unit circle are one and so i've proved the inequality here absolute values are 
never negative er suppose now that alpha is an algebraic integer with er say 
min- Q alpha 
the minimum polynomial that will be X to the oh we don't know what the 
coefficient is M plus A-one- X-to-the-M- minus-one plus er A-M with the A-Is in 
Z because we're supposing that it's an algebraic integer assume that alpha is 
strictly less than one i'll be done if i can show that this implies alpha 
equals nought by one of the facts i proved well this is er fourteen-point-eight 
fact one used here that's just the result that i have on the fourth line there 
[sniff] and by fact three we know that the conjugates of alpha where the omega-
one up to omega-D are conjugates of the roots of unity omega-one up to omega-D 
and are 
therefore again roots of unity i guess i forgot to divide out by D and this is 
all divided by D these are roots of unity their absolute value is one and so 
the conjugates also have absolute value equal to er or less than or equal to er 
er sorry the conjugates have absolute value equal to one therefore er er this 
conjugate has absolute value at most one now er the conjugates of alpha are the 
roots of that polynomial up there and as we know the product of the roots of a 
polynomial are plus or minus the constant term if you factorize the polynomial 
into linear factors the constant term is the product of the roots and that is 
of course an integer now the absolute value of the 
product of the conjugate is equal to the er things like that we made the 
assumption that alpha had absolute value strictly less than one so because of 
that last line there the absolute value of this is strictly less than one and 
equals the absolute value of A-N and that's a natural number now can anyone 
tell me a natural number whose absolute value is strictly less than one any 
volunteers you're on camera so you needn't be shy 
sm0942: is zero considered 
nm0941: yeah for the purposes of er this course zero is a natural number 
otherwise this would not make sense therefore A-N is zero and if 
the constant term of an irreducible polynomial is zero then X is a factor of it 
but it's irreducible therefore the polynomial is X itself and therefore er the 
root we have a unique root namely alpha which is zero so we took a number alpha 
of the given form which was strictly less than one in absolute value and we 
found it was zero so if we have an algebraic integer it's either one or zero in 
absolute value so that's a a bit of preparation and now the next result is 
right at the core of the proof of Burnside's theorem it's the really clever bit 
so this is the bit you have to watch carefully and make sure that i've got 
nothing up my sleeve so we'll call this one sixteen-point-two and it's 
sufficiently important to be called a theorem i think and it was due to 
Burnside er around 
nineteen-hundred long time ago they were clever even then these guys these 
group theorists er now what does it say er suppose G that's a finite group G 
has a conjugacy class with P-to-the-R elements so that stands for the conjugacy 
class so the number of conjugates is P-to-the-R for some prime P and some 
natural number R greater than or equal to one notice straight away this 
hypothesis tells me that the group is not abelian because in a c-, an abelian 
group all the conjugacy class c-, classes contain one element now we need a 
conclusion we've got a hypothesis the conclusion is that G is not a simple 
group and the definition of a simple group is one with exactly two normal 
subgroups 
it has a proper non-trivial if i can just squeeze it in here so that you can 
read it normal subgroup i've got to find some proper normal subgroup which is 
bigger than one i'm going to use just sixteen-point-one and the first thing i 
do is to [sniff] apply so this is the proof coming up very delicate proof i 
apply the orthogonality relationship apply it will be a column orthogonality 
relation to column one that's the one that has the degrees of the character in 
it corresponding to the conjugacy class of the identity element and 
the column of G G is the representative of the class with P-to-the-R and what 
do i get well nought we know the formula by now the inner product is er zero 
and er i have the sum over the irreducible characters running down the column 
of the character table of er well i can it's it's symmetric so i can write er 
chi- G chi-one-bar but chi-one is a natural number so chi-one-bar is chi-one 
and i i write this as er one corresponding to the trivial character plus the 
sum from I-equals-two up to the class number K of 
er g-, chi- I-G chi- I- one so in this sum i've taken away the trivial 
character chi-one and i've listed all the other irreducible characters chi-two 
up to chi-K and we know the number in all is little-K the class number of the 
group [sniff] so this is all it looks tricky but it's it's it's really rather 
nice so i d-, i divide by P because er well you'll see why i divide by P in a 
minute take this over to the other side divide by P and er what do i get i 
write it like that the sum from K-equals-two to K remember these are all non-
trivial irreducible characters of this product is minus-one-upon-P P's a prime 
so the right-hand side is therefore not an algebraic integer because the only 
rationals that are algebraic integers are the so-called rational 
integers the ordinary integers themselves 
sm0943: 
nm0941: have i got 
sm0943: 
nm0941: er ah thank you [laughter] that really threw you that one i can see 
[laughter] so the right-hand side not an algebraic integer surprise surprise 
therefore the left-hand side is not an algebraic integer now you have to 
remember a fact i proved no it's a fact i didn't prove in fact namely that the 
algebraic integers form a ring and therefore in particular er sums of products 
of 
algebraic integers are algebraic integers and on the left-hand side here i have 
er a sum of products so this is an algebraic integer here and if the sum of 
these products is to be an al-, is is not to be an algebraic integer at least 
one of these things must not be an algebraic integer there exists some I 
between two and K a-, and of course er yeah it's no good having this zero 
because that won't kill things off and er chi-I-G not equal to zero such that 
chi-I-one divided by P is again not an algebraic integer if all of these for 
non-zero values of that were algebraic integers then the left-hand side would 
be an algebraic integer that would be a contradiction in other words P does not 
divide the degree of this character chi-I if it did then this would be an 
integer therefore an algebraic integer at this point you cast your mind back to 
first year foundations did any of you here do it with me oh faithful that you 
are [laughter] still hanging in there we did something called er er a 
consequence of the Euclidean algorithm then and er you see what is yeah P is 
the power whoops P is the power of the prime for which G belongs to a conjugacy 
class with P-to-the-R elements er hence P and the number of elements in this 
conjugacy class sorry i should say there chi-of-one chi-I-of-one and this which 
is P-to-the-R but if you remember the other formula for the number of elements 
in a conjugacy class it's 
precisely the order of G divided by the order of the centralizer that was a 
consequence of the orbit-stabilizer theorem so these two numbers are coprime by 
the Euclidean algorithm we know there exist integers A and B such that the 
highest common factor of these two which is one is A times the order of G 
divided by the order of the centralizer of little-G er plus B times what do we 
have chi of chi-I-of-one a degree this is equal to the highest common factor 
two coprime integers you can always find A and B so that equation holds okay 
now we slightly manipulate this equation er multiply by chi 
sm0944: excuse me on the subscript for the centralizer 
nm0941: oh yeah have i got it wrong 
sm0944: you've got I and G which one is it 
nm0941: er oh it's j-, it's big-G it's a centralizer in the group yeah thank 
you multiply 
times by well what do we want er chi-I-of-G over chi- I-of-one to get now if 
things work out well the left-hand side is going to be in the form of something 
which we proved last time was an algebraic integer and then what have we got 
there we've got that's A times that and B times er chi-I-of-G upon chi-I-of-one 
this is equal to er have i got that right no i've forgotten to cancel out er so 
that's equal to one times chi-I-of- G over chi-I-of-one now i've put it in that 
form for the following reason look at the left-hand side er A is an integer 
that particular expression we showed last time in fifteen-point-one is an 
algebraic integer er we know the character values 
are algebraic integers hence the left-hand side is another algebraic integer 
being a sum of products of algebraic integers now look at the right-hand side 
and look at sixteen-point-one in your notes i've rubbed it out now by the first 
result we proved today [sniff] this is an algebraic integer and the result said 
this is either zero or one but the bottom here's a natural number the top here 
is non-zero by choice so it must be one so we manage to pick out an irreducible 
character [sniff] whose value on a non-trivial element coincides with the value 
on the identity which is the degree and we've already proved something about 
that er [sniff] er well l-, l-, before i go to case one let rho be the 
representation it's going to be an irreducible representation affording 
this irreducible character chi-I so chi-I is obtained by taking the trace of 
the matrices which are images under rho [sniff] so case one er rho is not 
faithful that means the kernel is non-trivial and the kernel can't be the whole 
of G because chi-I is not equal to chi-one [sniff] the only time the kernel is 
the whole group is when we have the trivial representation and the trivial 
character and so we're done in this case we have the kernel as a non-trivial 
proper normal subgroup are you following everyone following 
sm0945: 
nm0941: yeah question good 
sm0945: [laughter] 
nm0941: on this board 
sm0945: bo-, bottom right bottom right 
nm0941: in other words sorry 
sm0945: [laughter] 
nm0941: P does not divide 
sm0945: 
bottom right board 
nm0941: bo-, oh bottom right board okay okay [laughter] 
sm0945: you have in the brackets yeah 
nm0941: er sorry it's yeah it's that thing thanks yeah 
sm0945: 
nm0941: sorry that's exactly what fifteen-point-one says okay that that thing 
this is an integer that's an algebraic integer that's an integer that's an 
algebraic integer there for the whole of this [sniff] [sniff] so what's the 
other case er oh i'm proving that so i won't rub it out well the other case is 
where the the representation is faithful the kernel's one and so in this case 
the image of rho is isomorphic to G itself well if you go back to eight-point-
four it's a long 
time ago now but you may have a distant recollection 
sm0946: 
nm0941: mm-hmm 
sm0946: er you put er chi-I not equal to chi-one 
nm0941: chi-I equals er not sorry not equal to chi-one yeah you're r-, really 
sharp today thank you i'm not so sharp but you are er sorry about that er if 
it's faithful well by if you look back eight-point-four- A er the the statement 
there was that er rho-of-G i-, i-, i-, it's this condition star here using star 
that was the hypothesis of eight-point-four-A it said in this condition then 
the representation affording chi-I represents G as a scalar matrix so hence rho-
of-G because scalar matrices commute with all matrices that are D by D er 
that's 
in the centre of rho-of-G and it's not hard to see that that that's the same 
since rho is an isomorphism it's the image of the centre of G so apply the 
inverse map for rho since it's bijective [sniff] and that says that the group 
has non-trivial centre [sniff] now we saw at the very beginning that the 
hypothesis implied that G was not abelian er Z-G is a proper Z-G is only G if 
the group's abelian is a proper non-trivial normal subgroup okay well take a 
deep breath that was quite a bit of hard work you've done very well to keep 
your attention going i've got i just want to give you a short announcement 
about er ooh what bits of paper you're likely to get today you'll get something 
called Aims Objectives and Syllabus you may find it helpful to 
see what i was trying to do and what i've covered it's certainly er de rigueur 
to distribute one of these before the end of the course er on Friday i hope er 
very much earnestly to have had time to write you at least one mock exam paper 
so if you can't get along on Friday because Christmas calls then as soon as 
i've done it i'll stick some copies in the filing cabinet in the atrium yeah 
sm0947: er with respect to that you n-, you not recommend using 
nm0941: er i've had a look at last year's that namex set er yeah there are some 
bits and pieces there i mean what i would advise you to do have a look at it 
see if it figures he's more or less covered about half of what i've covered but 
from a different emphasis i've put more emphasis on character tables so you can 
make sense of some of the stuff there do what you can and i'll produce er my 
own mock exam and i will also put up the solutions next term on maths stuff so 
if you 
h-, are struggling with those questions by a couple of weeks into the term you 
should find the solutions on maths stuff so i i do i have got a few more things 
to do but i'll just send these round while er while i clean the boards would 
it's a bit would you mind just going up this this row 
sm0948: excuse me 
nm0941: mm 
sm0948: do you know when the test er when the 
nm0941: exam it's in April i think 
sm0948: no no no 
nm0941: okay in the last few minutes i want to just prove something which i 
think you may have seen already but i'll just use it as revision if you have a 
finite group er whose order is the power of some prime then the centre if it's 
if it's a a non-trivial group then the 
centre of G is also bigger than one well G is a union of conjugacy classes so 
when you add them all up you get the number of elements in the group is P-to-
the-M this is the order of G and then this is one-to-the-G G-two-to-the-G where 
one G-two G-three and so on are just representatives of the conjugacy classes 
now the number of conjugates are the identity is one and the other conjugates 
are P-to-the-M-two plus P-to-the-M-three plus so on because the number of 
elements in a conjugacy class is the index of the centralizer of an element so 
it divides the order of the group so the left-hand side is congruent to zero- 
modulo-P if M-two M-three and so on all bigger than one then the right-hand 
side is congruent to one-mod-P but that's nonsense because the left-hand side 
is congruent to zero-mod-P and you can't have that when P is the prime and so 
that's a contradiction there's my favourite little sign for a contradiction so 
some M-I i should say if they're greater than or equal to one i guess yep 
[sniff] so they can't all be greater than or equal to one so some M-I is zero 
then m-, mod- G-I-to-the-G is one P-to-the-zero which is one hence 
sm0949: you've got contradiction 
sm0950: that little lightning bolt's the 
nm0941: that is a contradiction right yeah i might just as well have written it 
out 
sm0949: 
nm0941: [laugh] 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 contradictory than that can you [laughter] i mean life would collapse er 
so what does it mean to say there's one conjugate of G-I it means whenever i 
conjugate by anything in G it stays the same that means it commutes with 
everything so it's in the centre of G it was not the identity because we 
separated the identity off and that gives me the result good well we're in g-, 
we're we're in pretty good shape for our climax on Friday so i look forward to 
seeing you then