# Isaacs Character Theory - exercise 4.11

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?

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.

• 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? Nov 26, 2022 at 16:00
• 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. Nov 26, 2022 at 19:08
• 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! Nov 27, 2022 at 3:21