Proving the Thompson Transfer Lemma

Question

Let $G$ be a finite group of even order $n=2^kr$, $T$ a Sylow-$2$ subgroup of $G$, and $M$ an index $2$ subgroup of $T$. I want to show that if $G$ has no subgroup of index $2$, then every element $x$ of order $2$ is conjugate to an element of $M$.

Using the transfer homomorphism, I get that $$Ver(x)=\prod g^{-1}xg \mod T'$$ is an element of $T/T'$, where the product is taken over representatives of the left cosets of $T$ that are fixed under left multiplication by $x$. Now, I want to show that if $G$ has no subgroup of index $2$, then this product is in fact in $M$, which would allow me to conclude that one of the factors is in $M$, since $x$ fixes an odd number of cosets and thus the product has an odd number of terms. But I'm not sure how to show this.

I know that if $G$ doesn't have an index $2$ subgroup, then all the permutations of its left regular representation are even. Since in the LRR, an element of order $m$ will consist of $\dfrac{n}{m}$ $m$-cycles, all the permutations being even implies that there is no element of order $2^k$, which means that $T$ is not cyclic. But I don't see how to use this information to conclude that $Ver(x)\in M/T'$.

Answer 1

Thompson's transfer lemma says important things about transfer, but actually the technique is more similar to Cayley's $2$-nilpotent criterion alluded to in your last paragraph.

Main hint, here is how you use the fixed points as in Cayley's theorem:

Consider the action of $G$ on the cosets $\{ Mg : g \in G \}$ of $M$. For an element $t$ of $G$ we have $(Mg)t = M(gt)$, by definition. If $t$ has a fixed point, then $(Mg)t = M(gt) = Mg$, so $gtg^{-1} \in M$ and $t$ has a conjugate in $M$. Conversely, if $gtg^{-1} \in M$, then $(Mg)t = Mg$ is a fixed point of $t$.

Here is the detailed conclusion (basically a repeat of Cayley's argument):

Now the hypothesis that $M$ is a maximal subgroup of $T$, a Sylow 2-subgroup of $G$ tells us that $[G:M]=4k+2$ for some non-negative integer $k$. If $t$ is an element of order $2$, then the action of $t$ on the cosets of $M$ is the product of some some transpositions. If $t$ has no fixed points (equivalently, if $t$ is not conjugate to an element of $M$), then it is the product of $[G:M]/2 = 2k+1$ transpositions, so is an odd permutation. Hence $t$ is not contained in the pre-image $N$ of the alternating subgroup of $\operatorname{Sym}(G/M)$. Hence $1 \neq [G:N] \leq [\operatorname{Sym}(G/M):\operatorname{Alt}(G/M)]=2$ leaving only $[G:N]=2$ as a possibility.

This does mean that $t$ is not an element of the kernel of the transfer of $G$ into $T$ (by the focal subgroup lemma), hence the name, but the actual transfer map is not used in proving that $N$ exists (and $t \notin N$).

Sometimes the details of $N$ can be discarded to get the simpler statement: if $M$ is a maximal subgroup of Sylow 2-subgroup of $G$, and $t$ is an element of order 2 contained in no conjugate of $M$, then $t$ is not contained in $O^2(G)$ (or even $E^2(G) \geq A^2(G) \geq O^2(G)$ if you are into those subgroups, the smallest normal subgroup with elementary abelian $2$-group, abelian $2$-group, or just plain $2$-group quotients).