If the commutator of a finite group has order $2$, then the order of the group is divisible by $8$ Prove that if $|G| < \infty$ and $|G'| = 2$ then $|G|$ is divisible by $8$.
Thoughts. $A \simeq G / G'$ is abelian and $G' \simeq \mathbb{Z}_2$. Since $G' \subset G$ then at least $|G| \vdots 2$. Now I'm stuck.
 A: In general, if $N$ is a normal subgroup order order $2$ of a group $G$, then $N \subseteq Z(G)$, the center of $G$. So $G' \subseteq Z(G)$.Let $P \in Syl_2(G)$. Note that $G' \subseteq P$, hence $P$ is even normal. A well-known theorem (proved with transfer) says that $P \cap G' \cap Z(G) \subseteq P'$. It follows that $G'\subseteq P'$, and of course $P' \subseteq G'$. So $|P'|=2$, and $P$ is non-abelian and must have at least order $8$.
A: Let me provide another (longer) proof, which is maybe more accessible. Firstly, if $N \unlhd G$, then one defines $[N,G]:=\langle [n,g]: n \in N, g \in G \rangle$, the subgroup generated by all commutators with one variable in $N$. Note that $[N,G]=[G,N]$ and that $[N,G]$ is a normal subgroup of $G$ contained in $N \cap G’$. In fact $[N,G]$ is the smallest normal subgroup $M$ of $G$, such that $N$ becomes central in the quotient $G/M$. We need a lemma.
Lemma Let $G$ be a group, with $G=HN$, where $H$ a subgroup and $N$ is a normal subgroup. Then $G’=H’[N,G]$.
Proof Since $H’ \subseteq G’$ and $[N,G] \subseteq G’$, it follows that $ H’[N,G] \subseteq G’$. Let us write $\overline{.}$ for reduction modulo $[N,G]$. Let $x=hm$ and $y=kn$, with $h,k \in H$ and $m,n \in N$. Let us calculate $[x,y]$ in $G/[N,G]$: $\overline{[x,y]}=[\overline{x},\overline{y}]=[\overline{hm},\overline{kn}]=[\overline{h}\overline{m},\overline{k}\overline{n}]=$ (here we use that $\overline{N} \subseteq Z(\overline{G})$) $=[\overline{h},\overline{k}]=\overline{[h,k]}$. This implies that $[x,y]=[h,k]g$, for some $g \in [N,G]$, whence $G’ \subseteq H’[N,G]$. $\square$
Now assume that $G$ is a group with $|G’|=2$. Let $P \in Syl_2(G)$. Then $G’ \subseteq P$ ($G’$ being a $2$-group is a subgroup of a conjugate of $P$ according to Sylow, but since $G’$ is normal, it is a subgroup of every Sylow $2$-subgroup!). But then $P$ is normal (in general if a subgroup contains the commutator subgroup, then it is normal). Now we apply the Schur-Zassenhaus Theorem (see for example M.I. Isaacs’ Finite Group Theory, chapter $3$): $G$ splits over $P$, that is $G=HP$, for a certain subgroup $H$, with $H \cap P=1$. Note that $G/P \cong H$ is abelian, since $G’ \subseteq P$. Now let us apply the lemma: $G’=H’[P,G]=[P,G] \supseteq [P,P]=P’$. We conclude $G’=P’ \cong C_2$ and then $P$ must be at least of order $8$. Or, $P'=1$, that is, $P$ is abelian. We have to rule out the last case.
So let $P$ be abelian. We can assume that $|P|$ is $2$ or $4$, otherwise we are done. Now $P=N_G(P)$ and $N_G(P)/C_G(P)$ embeds homomorphically into Aut$(P)$. $P$ is abelian, so $P \subseteq C_G(P)$, which means index$[G:C_G(P)]$ must be odd, since $P$ is a Sylow $2$-subgroup. If $|P|=2$, then $P \subseteq Z(G)$ and we would have $G=H \times P$ and $G$ is abelian, a contradiction. If $P \cong C_4$, then Aut$(P) \cong C_2$, whence $G=C_G(P)$, meaning $P$ being central, again a contradiction as in the previous case. So $P \cong C_2 \times C_2$. Put $K=HG'$. This subgroup has index $2$ in $G$, and index$[K:H]=2$. So $H$ is normal in $K$ and because $|H|$ is odd, $H$ is even characteristic in $K$. This means $H char K \lhd G$, so $H \lhd G$. This implies again $G=H \times P$ and $G$ being abelian, which is the final contradiction.
Finally note that proving the Schur-Zassenhaus Theorem requires an application of the group theoretic transfer.
In general, one can prove the following along the same lines:
If $G$ is a finite group with $|G'|=p$, the smallest prime dividing the order of $G$, then $p^3$ divides $|G|$.
