Commutator subgroup problem Let $G$ be a group and let $H$ and $K$ be subgroups of $G$. The commutator subgroup $[H,K]$ is defined as the smallest subgroup containing all elements of the form $hkh^{−1}k^{−1}$, where $h \in H$ and $k \in K$. Pick out the true statement(s):
 1. If $H$ and $K$ are normal subgroups, then $[H,K]$ is a normal subgroup.
 2. If $H$ and $K$ are characteristic subgroups, then $[H,K]$ is a characteristic subgroup.
 3. $[G,G]$ is normal in $G$ and $G/[G,G]$ is abelian.
 A: All three statements are true.
For part (a), let $g \in G, hkh^{-1}k^{-1}\in (H,K)$. Then $ghkh^{-1}k^{-1}g^{-1}=(gh^{-1}g^{-1})(gkg^{-1})(gh^{-1}g^{-1})(gk^{-1}g^{-1})=h_2k_2h_2^{-1}k_2^{-1}$ for some $h_2,k_2 \in H,K$. Thus $(H,K)$ is normal.
For part (b), we have that for all $\phi \in Aut(G), \phi(H)=H$ and $\phi(K)=K$. If $hkh^{-1}k^{-1} \in (H,K)$, then $\phi(hkh^{-1}k^{-1})=\phi(h)\phi(k)\phi(h^{-1})\phi(k^{-1})=h_2k_2h_2^{-1}k_2^{-1}$ for some $h_2,k_2 \in H,K$. This shows $(H,K)$ is invariant under any automorphism, hence it is a characteristic subgroup.
The first part of (c) follows directly from (a), since $G$ is normal in itself. To see the second part, realize that when you make a quotient group, you are essentially setting everything in the "denominator" equal to the identity. In this case, that means that for all $g_1,g_2 \in G$, we have $g_1g_2g_1^{-1}g_2^{-1}=1$, that is, $g_1g_2=g_2g_1$, hence the quotient group is abelian.
A: Basic Hints:
(1) A subgroup $\,H\leq G\,$ is normal iff $\,[H,G]\leq H\,$ . Thus, for example, if we have 
$$H,K\triangleleft G\Longrightarrow [H:K]\leq H\cap K$$
(2) For any endomorphism $\,\phi:G\to G\,\,\,,\,\,\phi[x,y]=[\phi(x),\phi(y)]\,\,\,,\,\,\forall\,x,y\in G$
