Burnside's theorem

Proof

The following proof — using more background than Burnside's — is by contradiction. Let ${\displaystyle p^a{q}^b}$ be the smallest product of two prime powers, such that there is a non-solvable group ${\displaystyle G}$ whose order is equal to this number.

• ${\displaystyle G}$ is a simple group with trivial center and ${\displaystyle a}$ is not zero.

If ${\displaystyle G}$ had a nontrivial proper normal subgroup ${\displaystyle H}$, then (because of the minimality of ${\displaystyle G}$), ${\displaystyle H}$ and ${\displaystyle G/H}$ would be solvable, so ${\displaystyle G}$ as well, which would contradict our assumption. So ${\displaystyle G}$ is simple.

If ${\displaystyle a}$ were zero, ${\displaystyle G}$ would be a finite q-group, hence nilpotent, and therefore solvable.

Similarly, ${\displaystyle G}$ cannot be abelian, otherwise it would be solvable. As ${\displaystyle G}$ is simple, its center must therefore be trivial.

• There is an element ${\displaystyle g}$ of ${\displaystyle G}$ which has ${\displaystyle q^d}$ conjugates, for some ${\displaystyle d>0}$.

By the first statement of Sylow's theorem, ${\displaystyle G}$ has a subgroup ${\displaystyle S}$ of order ${\displaystyle p^a}$. Because ${\displaystyle S}$ is a nontrivial ${\displaystyle p}$-group, its center ${\displaystyle Z(S)}$ is nontrivial. Fix a nontrivial element ${\displaystyle g\in Z(S)}$. The number of conjugates of ${\displaystyle g}$ is equal to the index of its centralizer ${\displaystyle G_g}$, which divides the index ${\displaystyle q^b}$ of ${\displaystyle S}$ (because ${\displaystyle S}$ is a subgroup of ${\displaystyle G_g}$). Hence this number is of the form ${\displaystyle q^d}$. Moreover, the integer ${\displaystyle d}$ is strictly positive, since ${\displaystyle g}$ is nontrivial and therefore not central in ${\displaystyle G}$.

• There exists a nontrivial irreducible representation ${\displaystyle \rho }$ with character ${\displaystyle \chi }$, such that its dimension ${\displaystyle n}$ is not divisible by ${\displaystyle q}$ and the complex number ${\displaystyle \chi (g)}$ is not zero.

Let ${\displaystyle ({\chi }_i{)}_{1\le i\le h}}$ be the family of irreducible characters of ${\displaystyle G}$ over ${\displaystyle \mathbb{C}}$ (here ${\displaystyle {\chi }_1}$ denotes the trivial character). Because ${\displaystyle g}$ is not in the same conjugacy class as 1, the orthogonality relation for the columns of the group's character table gives:

${\displaystyle 0=\sum _{i=1}^h{\chi }_i(1){\chi }_i(g)=1+\sum _{i=2}^h{\chi }_i(1){\chi }_i(g).}$

Now the ${\displaystyle {\chi }_i(g)}$ are algebraic integers, because they are sums of roots of unity. If all the nontrivial irreducible characters which don't vanish at ${\displaystyle g}$ take a value divisible by ${\displaystyle q}$ at 1, we deduce that

${\displaystyle -\frac{1}{q}=\sum _{i\ge 2,{\chi }_i(g)\ne 0}\frac{{\chi }_i(1)}{q}{\chi }_i(g)}$

is an algebraic integer (since it is a sum of integer multiples of algebraic integers), which is absurd. This proves the statement.

• The complex number ${\displaystyle q^d\chi (g)/n}$ is an algebraic integer.

The set of integer-valued class functions on ${\displaystyle G}$, ${\displaystyle Z(\mathbb{Z}[G])}$, is a commutative ring, finitely generated over ${\displaystyle \mathbb{Z}}$. All of its elements are thus integral over ${\displaystyle \mathbb{Z}}$, in particular the mapping ${\displaystyle u}$ which takes the value 1 on the conjugacy class of ${\displaystyle g}$ and 0 elsewhere.

The mapping ${\displaystyle A:Z(\mathbb{Z}[G])\to \mathrm{End}({\mathbb{C}}^n)}$ which sends a class function ${\displaystyle f}$ to

${\displaystyle \sum _{s\in G}f(s)\rho (s)}$

is a ring homomorphism. Because ${\displaystyle \rho (s{)}^{-1}A(u)\rho (s)=A(u)}$ for all ${\displaystyle s}$, Schur's lemma implies that ${\displaystyle A(u)}$ is a homothety ${\displaystyle \lambda I_n}$. Its trace ${\displaystyle n\lambda }$ is equal to

${\displaystyle \sum _{s\in G}f(s)\chi (s)=q^d\chi (g).}$

Because the homothety ${\displaystyle \lambda I_n}$ is the homomorphic image of an integral element, this proves that the complex number ${\displaystyle \lambda =q^d\chi (g)/n}$ is an algebraic integer.

• The complex number ${\displaystyle \chi (g)/n}$ is an algebraic integer.

Since ${\displaystyle q}$ is relatively prime to ${\displaystyle n}$, by Bézout's identity there are two integers ${\displaystyle x}$ and ${\displaystyle y}$ such that:

${\displaystyle x{q}^d+yn=1\text{therefore}\frac{\chi (g)}{n}=x\frac{q^d\chi (g)}{n}+y\chi (g).}$

Because a linear combination with integer coefficients of algebraic integers is again an algebraic integer, this proves the statement.

• The image of ${\displaystyle g}$, under the representation ${\displaystyle \rho }$, is a homothety.

Let ${\displaystyle \zeta }$ be the complex number ${\displaystyle \chi (g)/n}$. It is an algebraic integer, so its norm ${\displaystyle N(\zeta )}$ (i.e. the product of its conjugates, that is the roots of its minimal polynomial over ${\displaystyle \mathbb{Q}}$) is a nonzero integer. Now ${\displaystyle \zeta }$ is the average of roots of unity (the eigenvalues of ${\displaystyle \rho (g)}$), hence so are its conjugates, so they all have an absolute value less than or equal to 1. Because the absolute value of their product ${\displaystyle N(\zeta )}$ is greater than or equal to 1, their absolute value must all be 1, in particular ${\displaystyle \zeta }$, which means that the eigenvalues of ${\displaystyle \rho (g)}$ are all equal, so ${\displaystyle \rho (g)}$ is a homothety.

• Conclusion

Let ${\displaystyle N}$ be the kernel of ${\displaystyle \rho }$. The homothety ${\displaystyle \rho (g)}$ is central in ${\displaystyle \mathrm{Im}(\rho )}$ (which is canonically isomorphic to ${\displaystyle G/N}$), whereas ${\displaystyle g}$ is not central in ${\displaystyle G}$. Consequently, the normal subgroup ${\displaystyle N}$ of the simple group ${\displaystyle G}$ is nontrivial, hence it is equal to ${\displaystyle G}$, which contradicts the fact that ${\displaystyle \rho }$ is a nontrivial representation.

This contradiction proves the theorem.