TITLE: Is a non-trivial finite perfect group of order 4n?
QUESTION [6 upvotes]: A finite group $G$ is perfect if $G = G^{(1)} := \langle [G,G] \rangle$, or equivalently, if any $1$-dimensional complex representation is trivial.
Question: Is a non-trivial finite perfect group of order a multiple of $4$?
Remark: It's true for $\vert G \vert \le 10^6$, by checking with GAP.

REPLY [3 votes]: Out of interest, here is another approach to prove that a finite group $G$ which has a cyclic Sylow $2$-subgroup $S$ necessarily has a normal $2$-complement. By induction, it suffices to exhibit a (normal) subgroup of index $2$.
   Consider the regular representation $\rho_{G}$ of $G$. This is a representation of $G$ by permutation matrices, so $\delta(x) = {\rm det} \rho_{G}(x) = \pm 1$ for all $x \in G$. Hence ${\rm ker} \delta$ is a normal subgroup of $G$ of index dividing $2$. Hence it suffices to find an element $x \in G$ with $\delta(x) = -1.$
Let $s$ be a generator of $S$. Note that ${\rm Res}^{G}_{S}(\rho_{G}) = [G:S]\rho_{S}$, so that $\delta(s) = \phi(s)^{[G:S]}$, where $\phi(s) = {\rm det} \rho_{S}(s).$ Since $[G:S]$ is odd, it suffices to prove that $\phi(s)= -1$.
Now the eigenvalues of $\rho_{S}(s)$ are all the complex (but not necessarily primitive) $|S|$-th roots of unity, each occurring with multiplicity $1$. 
However, apart from $1$ and $-1$, all other such roots of unity occur in complex conjugate pairs, and each conjugate pair contributes $1$ to the determinant.
Hence $\phi(s) = 1 \times -1 = -1$, as required.