Second derived group Let $G'$ be the commutator group of the group $G$, and $G''$ the commutator group of $G'$. Show that if $G''$ is cyclic then $G''\subseteq Z(G')$, where $Z(G')$ is the centre of $G'$).
This can't be that difficult - it is half an exercise in Martin Isaac's Algebra (and most exercises have solutions that are only a few lines long). But I have wasted an inordinate amount of time failing to find the solution, so I must be missing something obvious.
 A: Hint: $G/C_G(G$''$)$ can be homomorphically embedded in Aut$(G$''$)$, which is abelian.
A: For any finite cyclic group $H$ with generator $t$, an automorphism of $H$ is completely determined by its effect on $t$ (which it must take to $t^k$ for some $k$ prime to the order of $t$). Hence Aut($H$) is abelian. 
Next we show that if $H$ is any subgroup of $G$, then $N_G(H)/C_G(H)$ is isomorphic to a subgroup of Aut($H$). Define $\theta:N_G(H)\to $Aut($H$) by $\theta(x)(h)=x^{-1}hx$. Evidently it has kernel $C_G(H)$, so the claim follows.
Applying these two observations to $H=G''$, where $G''$ is cyclic, we conclude that $N_G(G'')/C_G(G'')$ is isomorphic to a subgroup of an abelian group and hence abelian.
Finally, repeated application of $(xy)^g=x^gy^g$ shows that $[x,y]^g=[x^g,y^g]$, and repeated application of that shows that if $k\in G''$ and $g\in G$, then $k^g\in G''$ and so $N_G(G'')=G$.
Take any $x,y\in G$. Then $xyC_G(G'')=yxC_G(G'')$ and hence $x^{-1}y^{-1}xyC_G(G'')=C_G(G'')$, so $[x,y]\in C_G(G'')$ and hence $G'\subseteq C_G(G'')$. In other words, any element of $G'$ commutes with any element of $G''$ and so $G''\subseteq Z(G')$ 
