$G\times H\cong G$ with $H$ non-trivial Aluffi II.3.4 asks the reader to find groups $G$ and $H$ such that $G\times H\cong G$ but $H$ is not trivial. I believe I have found a solution, but would like someone to check. Also, is there a more elementary example? I tried mucking around with $S^1\times (\Bbb Z/2\Bbb Z)$, which feels like it should be isomorphic to $S^1$ (the unit circle under complex multiplication), but I kept getting confused.
My solution
Let $H$ be a non-trivial group and let $G=H^\omega$. Then $\phi\colon H\times G\to G$ defined by
$$\phi(h,g)_n=\begin{cases}
h &n=0\\
g_{n-1} &n\ne 0
\end{cases}$$
looks like a group isomorphism to me. It's obviously a bijective function, and \begin{align*}
\phi((h_1,g_1)(h_2,g_2))_n
&=\phi(h_1h_2,g_1g_2)_n \\
&=\begin{cases}
h_1h_2&n=0\\
(g_1g_2)_{n-1}&n\ne 0
\end{cases}\\
&=\phi(h_1,g_1)_n\phi(h_2,g_2)_n \\
&=(\phi(h_1,g_1)\phi(h_2,g_2))_n.
\end{align*}

A simple example would be the natural numbers (including zero) under the operation "bitwise exclusive or." This is isomorphic to $$G=\oplus_{i=0}^{\infty} \mathbb Z/\langle 2\rangle$$
