# Abstract Nonsense

## Burnside’s Theorem

Point of post: In this post we put together a lot of our rep theory to prove one of the fundamental (pure) group theoretic results amenable to the subject.

Motivation

In this post we finally use representation theory to prove something in pure group theory that is near impossible to do without representation theory. We have seen on our thread about solvable groups that every $p$-group is solvable. In this thread we prove Burnside’s Theorem an amazing generalization which says that every group of order $p^aq^b$ where $p$ and $q$ are primes. As a corollary we will be able to conclude that every non-abelian simple group is divisible by three distinct primes which, of course, will eliminate a respectable amount of group orders for analyzing simplicity. This is really one of the most beautiful applications of representation theory.

Propositions

Before we prove the actual theorem we shall need to prove a series of propositions which make the proof of Burnside’s theorem possible.

Proposition:

Before proving the full theorem we shall need to prove some lemmas which are interesting in and of themselves.

Proposition #1: Let $G$ be a finite group and $\chi\in\text{irr}(G)$. Suppose that there existed a conjugacy class $\mathcal{C}$ of $G$ such that $\left(\#\left(\mathcal{C}\right),\chi(e)\right)=1$ then for every $g\in\mathcal{C}$ either $\chi(g)=0$ or $g\in\mathbf{Z}(\chi)$ where $\mathbf{Z}(\chi)$ is the center of the character $\chi$.

Proof: Since $\left(\#\left(\mathcal{C}\right),\chi(e)\right)=1$ we have by Bezout’s identity there exists $m,n\in\mathbb{Z}$ such that $m\#\left(\mathcal{C}\right)+n\chi(e)=1$ and thus multiplying by $\chi(g)$ for $g\in\mathcal{C}$ gives $m\#\left(\mathcal{C}\right)\chi(g)+n\chi(e)\chi(g)=\chi(g)$ dividing both sides by $\chi(e)$ gives

$\displaystyle m\#\left(\mathcal{C}\right)\frac{\chi(g)}{\chi(e)}+n\chi(g)=\frac{\chi(g)}{\chi(e)}$

we know them from a previous theorem that $\displaystyle \frac{\chi(g)}{\chi(e)}\#\left(\mathcal{C}\right)$ is an algebraic integer and since $\chi(g) n$ is evidently an algebraic integer, and the algebraic integers $\mathbb{A}$ are a ring that $\displaystyle \alpha=\frac{\chi(g)}{\chi(e)}\in\mathbb{A}$.

$\text{ }$

Suppose then that $g\notin\mathbf{Z}(\chi)$, then by definition we have that $|\alpha|<1$. Let $m=|g|$ and let $k$ be the splitting field of $x^m-1\in\mathbb{Q}[x]$, and note then that necessarily $\alpha\in k$. Now, note that $\chi(g)$ is a sum of $\chi(e)^{\text{th}}$ roots of unity, and since any $\sigma\in\text{Gal}(k/\mathbb{Q})$ necessarily takes $\chi(e)^{\text{th}}$ roots of unity to other such roots of unity we may conclude that $\sigma(\chi(g))$ is a sum of such roots of unity for each $\sigma\in\text{Gal}(k/\mathbb{Q})$.  From this it follows that $|\sigma(\chi(g))|\leqslant \chi(e)$ and so $|\sigma(\alpha)|\leqslant 1$ for each $\sigma\in\text{Gal}(k/\mathbb{Q})$. Now from this we may conclude that

$\text{ }$

$\displaystyle \left|\prod_{\sigma\in\text{Gal}(k/\mathbb{Q})}\sigma(\alpha)\right|<1$

$\text{ }$

(since $|\alpha|<1$ by assumption) but note that since elements of $\text{Gal}(k/\mathbb{Q})$ necessarily take $\mathbb{A}$ into itself we must have that each $\sigma(\alpha)$ is in $\mathbb{A}$ and so their product, as above, is in $\mathbb{A}$. That said, since any $\sigma\in\text{Gal}(k/\mathbb{Q})$ just permutes the relevant roots of unity we clearly have that $\displaystyle \prod_{\sigma\in\text{Gal}(k/\mathbb{Q})}\sigma(\alpha)$ is in the fix field $k^{\text{Gal}(k/\mathbb{Q})}$. That said, since $k$ is the splitting field of the separable polynomial $x^m-1$ we know that $k/\mathbb{Q}$ is Galois and so $k^{\text{Gal}(k/\mathbb{Q})}=\mathbb{Q}$ and thus $\displaystyle \prod_{\sigma\in\text{Gal}(k/\mathbb{Q})}\sigma(\alpha)$ is rational. But, since the only rational algebraic integers are integers we know that this product is an integer, and since it’s modulus is less than one it must be zero. Thus, this tells us that $\sigma(\alpha)=0$ for some $\sigma\in\text{Gal}(k/\mathbb{Q})$ and so $\alpha=0$, so $\chi(g)=0$ as desired. $\blacksquare$

$\text{ }$

Proposition #2: Let $G$ be a non-abelian simple group. Then, $\{e\}$ is the only conjugacy class of $G$ which is a prime power.

Proof: Suppose that $g\in G$ is such that $\#\left(\mathcal{C}_g\right)=p^a>1$ where $\mathcal{C}_g$ is the conjugacy class containing $g$ and $p$ is prime. Note that for any $\chi\in\text{irr}(G)$ distinct from the trivial character we have that $\ker\chi=\bold{Z}(\chi)=\mathcal{Z}(G)=\{e\}$. The first of these is true since by simplicity and so $\ker\chi=\{e\}$ or $G$ and it must be $\{e\}$ otherwise a quick argument (considering that $\ker\chi$ is really the kernel of any character inducing $\chi$) shows that $\chi$ is the trivial character. The second is true since once again by simplicity $\bold{Z}(\chi)=\{e\}$ or $G$, and if $\bold{Z}(\chi)=G$ then using the fact that $\bold{Z}(\chi)/\ker\chi=\mathcal{Z}\left(G/\ker\chi\right)$ (since $\chi\in\text{irr}(G)$) and the fact that $\ker\chi=\{e\}$ so $\left|G/\ker\chi\right|=|G|$ we may conclude that $\mathcal{Z}(G)=G$ contradicting that $G$ is not abelian. Lastly, the fact that $\mathcal{Z}(G)=\{e\}$ is clear since otherwise $G$ is abelian, appealing to simplicity again of course.

So, if $p\nmid \chi(e)$ then by proposition #1 (since $g\ne e$ and so $g\notin\mathbf{Z}(\chi)$) we have that $\chi(g)=0$. So, using the second orthogonality relation we see that

$\displaystyle 0=\sum_{\alpha\in\widehat{G}}\chi^{(\alpha)}(e)\chi^{(\alpha)}(g)=1+\sum_{\alpha\in\widehat{G}\;\;\text{s.t. }p\mid\chi^{(\alpha)}(e)}\chi^{(\alpha)}(e)\chi^{(\alpha)}(g)$

So that

$\displaystyle -1=p\sum_{\alpha\in\widehat{G}\;\;\text{ s.t. }p\mid\chi^{(\alpha)}(e)}\frac{\chi^{(\alpha)}(e)}{p}\chi^{(\alpha)}(g)$

and since each $\displaystyle \frac{\chi^{(\alpha)}(e)}{p}\chi^{(\alpha)}(g)\in\mathbb{A}$ we may conclude from the above that $\displaystyle \frac{-1}{p}\in\mathbb{A}$. But, since this is a non-integer rational this is a contradiction. $\blacksquare$

Burnside’s Theorem

And finally the result:

Theorem (Burnside): Let $G$ be a finite group with $\left|G\right|=p^aq^b$ where $p$ and $q$ are primes. Then, $G$ is solvable.

Proof: We note that we can assume without loss of generality that $a,b\geqslant 1$ otherwise $G$ is a $p$-group and the conclusion follows from a previous theorem.

We first show that if $G$ is a simple group of order $p^aq^b$ where $p$ and $q$ are primes then it is abelian. Indeed, by Sylow’s first theorem we know that $G$ must have a subgroup of order $p^a$. If $G$ has only one of these such subgroups, then by Sylow’s second theorem we have that this subgroup is normal, and so $G$ has a non-trivial (via our opening remark) proper normal subgroup contradicting simplicity, so assume not. So, let $P$ be one of these Sylow $p$-subgroups.  By first principles we know that $\mathcal{Z}(P)$ is non-trivial. So, we may choose a non-trivial $g\in\mathcal{Z}(P)$. From the orbit stabilizer theorem we know that $\#\left(\mathcal{C}_g\right)=\left(G:\bold{C}_G(g)\right)$ where $\mathcal{C}_g$ is the conjugacy class containing $g$ and $\bold{C}_G(g)$ $g$‘s centralizer. It follows from Lagrange’s theorem that $q^b=\left(G:P\right)=\left(G:\bold{C}_G(g)\right)\left(\bold{C}_G(g):P\right)$ and thus $\left(G:\bold{C}_G(g)\right)=\#\left(\mathcal{C}_g\right)$ is a prime power. But, by proposition # 2 this implies that $G$ can’t be non-abelian and simple and since it is simple it follows that $G$ is abelian.

We now use this to prove the full result. We induct on $a+b$. If $a+b=1$ then $G$ is a cyclic group and thus trivially solvable. So, assume that every group of order at most $p^aq^b$ where $a+b=m$ is solvable and let $G$ be a group of order $p^aq^b$ where $a+b>m$. If $G$ is simple we’re done since by the first part of the proof $G$ is abelian and so trivially solvable. So, assume that $G$ is not simple. Then, $G$ has $\{e\}\triangleleft N\triangleleft G$. But, this implies that $N$ and $G/N$ are groups of order $p^aq^b$ where $a+b and thus by the induction hypothesis they themselves are solvable. But, by a prior theorem this implies that $G$ itself is solvable. Thus, the induction is complete and the conclusion follows. $\blacksquare$

References:

1. Isaacs, I. Martin. Character Theory of Finite Groups. New York: Academic, 1976. Print.

March 11, 2011 -

## 1 Comment »

1. […] very difficult (if not impossible) to prove without using representation theory– of course Burnside’s Theorem is the quintessential example. In this post we prove another such theorem, namely that for a finite […]

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