$H_1\times H_2 Let $H_1, H_2, G_1, G_2$ be groups. Clearly if $H_1<G_1$ and $H_2<G_2$, then $H_1\times H_2<G_1\times G_2$. 
I'm wondering if the converse statement is true. I'm quite sure it's not. Can you find a counterexample?

The subgroup $\;\langle (1,1)\rangle=\{(0,0)\,,\,(1,1)\} \le \Bbb Z_2\times\Bbb Z_2\;$ can be taken as $\;\langle (1,1)\rangle\times\{0\}\;$ , and the first factor isn't a subgroup of $\;\Bbb Z_2\;$ , of course...
