If $K$ is a normal subgroup of $G \times H$, which has only the identity in common with $G \times {e}$ and ${e} \times H$, then $K$ is abelian.

My attempt: take an element $x = (aba^{-1}b^{-1}, \alpha\beta\alpha^{-1}\beta^{-1}) \in [K, K]$. I know want to prove that either $aba^{-1}b^{-1}$ or $\alpha\beta\alpha^{-1}\beta^{-1}$ must be the identity, because that would mean that $x$ is the identity, and then $K$ is abelian. I tried this by contradiction: suppose $aba^{-1}b^{-1}$ and $\alpha\beta\alpha^{-1}\beta^{-1}$ are both not the identity. Since $K$ is a normal subgroup, $gK = Kg$ for any $g \in G \times H$. So in particular if $g = ((aba^{-1}b^{-1})^{-1}, (\alpha\beta\alpha^{-1}\beta^{-1})^{-1}) \in K$, then $gx = zg$ for some $z \in K$, meaning that $e = z((aba^{-1}b^{-1})^{-1}, (\alpha\beta\alpha^{-1}\beta^{-1})^{-1})$, so $z = x$, but I don't really see if/how this is a contradiction, or what else I can try.

(For posterity: this is exercise 15.7 from Groups and Symmetry by M.A. Armstrong)

For convenience, I'm going to write $e$ as identity element of all groups.

Step 1:

Consider the subgroup $K_G=\{(g,e)|(g,h)\in K$ for some $h\in H\}$.

Step 2:

As $K$ is normal in $G\times H$, $g^{-1}Kg=K$ for all $g\in K_G$.

Step 3:

Let $(a,b),(c,d)\in K$. By the above, $(a^{-1}ca,d)=(a,e)^{-1}(c,d)(a,e)\in K$, so $G\times e\ni (c^{-1}a^{-1}ca,e)=(c,d)^{-1}(a^{-1}ca,d)\in K$. Therefore $(c^{-1}a^{-1}ca,e)=e$ and $c^{-1}a^{-1}ca=e$. That is $ca=ac$. But by a similar argument $db=bd$, so $(c,d)(a,b)=(a,b)(c,d)$ and $K$ is abelian.

• I don't understand why $(c^{-1}a^{-1}ca,e)=(c,d)^{-1}(a^{-1}ca,d)$, could you please explain it? – user423841 Oct 7 '17 at 10:01
• $(c,d)^{-1}(a^{-1}ca,d)=(c^{-1},d^{-1})(a^{-1}ca,d)=(c^{-1}a^{-1}ca,d^{-1}d)=(c^{-1}a^{-1}ca,1)$. Does each step make sense? If not which one? – Robert Chamberlain Oct 7 '17 at 10:11

Theorem If $N \unlhd G_1 \times G_2$, then either $N \subseteq Z(G_1 \times G_2)$ or $N$ intersects one of the factors $G_1$ or $G_2$ non-trivially.

Proof. Assume that $N \cap (G_1 \times$ {$1$}$)$ = {$(1,1)$} = $N \cap$({$1$} $\times$ $G_2)$. We will show that N is contained in the center of $G_1 \times G_2$. Fix an arbitrary $(n_1,n_2) \in N$. Let $g_1 \in G_1$. Since $N$ is normal, $(g_1,1)^{-1}·(n_1,n_2)·(g_1,1) = (g_1^{-1}n_1g_1, n_2) \in N.$ Obviously also $(n_1,n_2)^{-1} = (n_1^{-1},n_2^{-1}) \in N$. It follows that the product $(n_1^{-1},n_2^{-1})·(g_1^{-1}n_1g_1, n_2) = (n_1^{-1}g_1^{-1}n_1g_1, 1) \in N$. However, $N$ intersects $G_1 \times$ {$1$} trivially, whence $n_1^{-1}g_1^{-1}n_1g_1 = 1$, that is, $n_1$ commutes with every $g_1 \in G_1$. Similarly, $n_2$ commutes with every $g_2 \in G_2$. This means that $(n_1,n_2) \in Z(G_1) \times Z(G_2) = Z(G_1 \times G_2)$. $\square$