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Suppose that $G$ is a group and that $n$ is a positive integer diving the order of $G$. Let $f_n(G)$ be the number of elements satisfying $x^n = 1$ in $G$. According to a theorem of Frobenius, then we have $f_n(G) \equiv 0 \mod{n}$. Hence if we have a family of groups and a formula for the number of solutions to $x^n = 1$, an application of Frobenius theorem proves that the formula is $\equiv 0 \mod{n}$.

One family of groups where we can find a formula are the symmetric and alternating groups, since the order of a permutation is determined by cycle structure. For example, let $p$ be a prime number and $n \geq p$ some integer. Then the number of elements satisfying $x^p = 1$ in $S_n$ is

$$\sum_{k = 1}^{\lfloor \frac{n}{p} \rfloor} \frac{n!}{p^k (n-pk)! k!} + 1 \equiv 0 \mod{p}$$

by Frobenius theorem. It does not seem immediately obvious without Frobenius theorem that the sum on the left should be $\equiv -1 \mod{p}$. Also, notice that $n = p$ gives us Wilson's theorem: $$(p-1)! \equiv -1 \mod{p}$$

More congruences can be found by calculating the number of elements of some order in the symmetric groups $S_n$ and alternating groups $A_n$.

My question is this:

Besides $S_n$ and $A_n$, for what other families of finite groups can we find congruences like the one above by applying Frobenius theorem?

Of course, if you feel like there's some noteworthy/interesting/useful congruence in $S_n$ to be found with Frobenius theorem, feel free to answer or comment with those too.

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As a somewhat related problem, perhaps you will be interested in this one (for which Mr. Gruber also provided an excellent answer): math.stackexchange.com/questions/219993/… –  Benjamin Dickman Jan 19 '13 at 22:01
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1 Answer

We can combine this with another theorem of Frobenius I recently used for interesting results.

Theorem. (Frobenius-Schur) If $t$ is the number of involutions in $G$, $$1+t=\sum_{\chi\in\text{Irr}(G)}a_\chi\chi(1)$$ where $a_\chi=0$ if $\chi\not= \overline{\chi}$ and $a_\chi=\pm 1$ otherwise. Furthermore, $\chi$ is afforded by a real representation if and only if $a_\chi=1$.

In the linked question, for example, I used this to bound the number of involutions of generalized dihedral groups with composite order by counting them explicity. You can use this technique to compute $f_2(G)$ for any group with an easy set of character degrees.

Let's compute $f_2(G)$ for $G=\text{GL}_n(\mathbb{F}_q)$, where $q$ is odd. Let $\mathcal{i}$ be an involution. Then every eigenvalue of $\mathcal{i}$ is $1$ or $-1$. Denoting the multiplicity of $-1$ in this diagonal matrix by $k$ (and thus the multiplicity of $1$ by $n-k$), we then have $$C_G(\mathcal{i})\cong \text{GL}_{n-k}(\mathbb{F}_q)\times \text{GL}_{k}(\mathbb{F}_q).$$ We know that $$|\text{GL}_d(\mathbb{F}_q)|=\prod_{m=0}^{d-1}\left(q^d-q^m\right),$$ so we can use that to compute the size of each conjugacy class of involutions: $$ \begin{eqnarray*}[G:C_G(\mathcal{i})]&=&\frac{|\text{GL}_n(\mathbb{F}_q)|}{|\text{GL}_{n-k}(\mathbb{F}_q)\times \text{GL}_{k}(\mathbb{F}_q)|}\\ &=&\frac{\prod_{m=0}^{n-1}\left(q^n-q^m\right)}{\prod_{m=0}^{n-k-1}\left(q^{n-k}-q^{m}\right)\prod_{m=0}^{k-1}\left(q^{k}-q^m\right)}\\ &=&q^{k(n-k)}\left(\frac{\prod_{m=1}^n(q^m-1)}{\prod_{m=1}^{k}(q^m-1)\prod_{m=1}^{n-k}(q^m-1)}\right)\\ &=&q^{k(n-k)}{n \choose k}_q,\end{eqnarray*}$$ where ${n \choose k}_q$ is a $q$-binomial. There is at least one involution with $k=m$ for $m=1$ to $n$ (namely, the diagonal matrix with that many $-1$s), so summing over each conjugacy class of involutions, we find $$\sum_{k=1}^nq^{k(n-k)}{n \choose k}_q=-1+\sum_{\chi\in \text{Irr}(\text{GL}_n(\mathbb{F}_q))}\chi(1)=-1+f_2(\text{GL}_n(\mathbb{F}_q)).$$ This is an appreciably interesting formula for $f_2(GL_n(\mathbb{F}_q))$, and additionally, by the theorem of Frobenius the OP provides, a pretty neat proof that $\sum_{k=0}^nq^{k(n-k)}{n \choose k}_q$ is always even.

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I have to admit I'm not familiar with some concepts used here (character theory, $q$-binomials), but this is exactly what I was looking for. Thanks for the answer. –  Mikko Korhonen Jan 20 '13 at 13:48
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