I am trying to prove the following statement from Isaac's "Character theory of finite groups":

Let $G$ be simple and let $S \in \operatorname{Syl_{2}}(G)$ be elementary abelian; $|S| = q.$ Suppose $S = C_{G}(x) \; \forall x \in S; x \neq 1.$ Show that $\nu_{2}(\chi) = 1 \; \forall \; \chi \in \operatorname{Irr}(G)$ and that $\chi(x) = \chi(1) - q$ for $\chi \neq 1_{G}$ and $x \in S$ with $x \neq 1$.

I have a few questions about this problem. I would like to say in advance that I wouldn't really like to apply the classification of simple groups to solve this problem.

I tried to start the proof like this:

Let $t$ be the number of involutions of the group $G$. Take an arbitrary involution $g\in S$. Then $$C_{G}(g) = S \Rightarrow t \geq \frac{|G|}{|C_{G}(g)|} = \frac{|G|}{q}.$$

It's not difficult to understand that in the group $G$ there are only involutions and elements of odd order.

Indeed, let $x$ be an element of even order unequal to $2$. Let $|x| = k$. Then $x^{\frac{k}{2}}$ is an involution and $C_{G}(x^{\frac{k}{2}})$ is an elementary Abelian Sylow $2$-subgroup of $G$. But $x\in C_{G}(x^{\frac{k}{2}})$ is a contradiction.

$\textbf{Question №1:}$ how to prove that the number of involutions are exactly $\frac{|G|}{q}$?

It's clear that it's enough, taking into account the arguments presented above, to count the various elements of all the Sylow $2$-subgroups. But as for me, it's quite difficult.

To understand why $\nu_{2}(\chi) = 1\; \forall\;\chi\in \operatorname{Irr}(G)$ it's enough to show that $\nu_{2}(\chi) > 0 \; \forall\;\chi\in \operatorname{Irr}(G).$

Let's say we proved that the number of involutions is exactly $\frac{|G|}{q}.$ Let $\chi \in \operatorname{Irr}(G).$ Then $$\nu_2(\chi) = \frac{1}{|G|}\sum\limits_{g \in G}\chi(g^{2}) = \frac{1}{|G|}(\frac{|G|}{q} + 1)\chi(1) + \frac{1}{|G|}\sum\limits_{g \in G; |g| > 2}\chi(g^{2}).$$

Since $G$ is a simple group, then $Z(\chi) = Z(G/\ker\chi)= Z(G) = 1.$ This means that only an unit element can correspond to a diagonal matrix.

$\textbf{Question №2:}$ How to prove using the above facts that $\nu_{2}(\chi) >0$?

Using the above facts, I tried to make an estimate for $\nu_2(\chi)$ (to be more precise for this expression: $\sum\limits_{g \in G; |g| > 2}\chi(g^{2})$). However, each time I got it too rough, because of which I could not prove that $\nu_{2}(\chi)>0$.

Any help?


1 Answer 1


For Question 1, to show that $t = |G|/q$, it is sufficient to show that all involutions in $G$ are conjugate. There is a hint on how to do that in the book.

If not, choose two non-conjugate involutions that lie in $x,x'$ in distinct Sylow $2$-subgroups $S$ and $S'$.

Since $x,x'$ are not conjugate, $xx'$ must have even order $2k$ for some $k$, and then $(xx')^k$ is an involution centralizing both $x$ and $x'$.

But the condition $C_G(x) = S$ for all $1 \ne x \in X$ forces $S$ and $S'$ to be disjoint, but $C_G(x') = S'$, so we have a contradiction.

For the second question, note that elements of odd order have a unique square root, so $$\sum_{g \in G,|g| > 2}\chi(g^2) = \sum_{g \in G,|g| > 2}\chi(g),$$ and for $\chi \ne 1_G$ we have $$\sum_{g \in G}\chi(g)=0.$$ Combining that with your equation for $\nu_2(\chi)$ gives $\chi(g) = \chi(1)-q,\,\chi(1)$, or $\chi(1)+q$ for $1 \ne g \in S$ when $\nu_2(\chi) = 1,\,0$, or $-1$, respectively.

But $\chi(g)=\chi(1)$ or $\chi(1)=q$ is not possible, so the result follows.

  • $\begingroup$ Thank you very much for your help! Sorry I didn't think of such simple steps. I would like to clarify one question. Was it possible to prove the conjugacy of involutions using the Sylow theorem that all Sylow subgroups are conjugate? $\endgroup$
    – Mr. Nobody
    Nov 26, 2022 at 16:00
  • $\begingroup$ No I don't think it follows from that. It is possible to prove it using Burnside's Transfer Theorem, but the method suggested in Isaacs' book, which is the one I used here, is simpler and more elementary. $\endgroup$
    – Derek Holt
    Nov 26, 2022 at 19:08
  • $\begingroup$ Yes, I completely agree with you. But I just wanted to know about the solution from the point of view of this approach. Thanks a lot! $\endgroup$
    – Mr. Nobody
    Nov 27, 2022 at 3:21

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