Is a non-trivial finite perfect group of order 4n? 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.
 A: 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.
A: Yes, if the $2$-Sylow subgroup is cyclic then the group has a normal $2$-complement, which is thus a proper normal subgroup with abelian quotient (being of order $2$ in the case where the order is divisible by $2$ but not by $4$).
The above statement follows from Burnside's transfer theorem together with the normalizer/centralizer theorem.
In fact, one can also see in the same way that if $4$ divides the order of the group and $8$ does not, then if the group is perfect it must have order divisible by $3$.
A bit more about the general case: Burnside's transfer theorem states that if the group has a $p$-Sylow subgroup which is central in its normalizer, then the group has a normal $p$-complement. Obviously, if a group has a normal $2$-complement then it is solvable (by Feit-Thompson) and hence not perfect.
To apply it in this situation, we note that if $P$ is the $2$-Sylow then what we need is to show that $N_G(P)/C_G(P)$ is trivial, and using that this is isomorphic to a subgroup of $\operatorname{Aut}(P)$, what we need to know something about is the order of this automorphism group.
For small orders of $P$, it is easy to give explicit orders for the automorphism groups, but in general, if one wants to make a statement of the form "if $2^m$ is the largest power of $2$ dividing $|G|$ and $G$ is perfect, then $|G|$ is divisible by one of the following primes" one would need to understand the set $X_n$ of odd primes dividing the orders of automorphism groups of groups of order $2^n$, and I don't think much is known about this in general, apart from the obvious fact that $X_n\subseteq X_{n+1}$ (though probably it should be quite possible to say something with the added assumption that the $2$-Sylow is abelian). 
