# $G \cong G \times H$ does not imply $H$ is trivial.

In Aluffi's Algebra: Chapter $0$ there is a question asking to give a counterexample to the claim

$G \cong G \times H$ implies $H$ is trivial.

I am looking for a hint. Obviously, at least one of $G$ or $H$ needs to be infinite. Doing something with $\mathbb{Z}$ seems to be the natural thing. I tried showing $\mathbb{Z} \cong \mathbb{Z} \times (\mathbb{Z}/2\mathbb{Z})$ by an interlacing evens and odds'' argument, but the "odd + odd" case killed my homomorphism...

Am I on the right track?

Thanks.

• How about $G$ group of finite sequences valued in $\mathbb{Z}$, ie $G = \oplus_{i = 0}^\infty \mathbb{Z}$ and $H = \mathbb{Z}$. – user174456 Sep 10 '14 at 3:40
• Your idea doesn't work. $\mathbb Z \times (\mathbb Z/2\mathbb Z)$ has torsion, while $\mathbb Z$ does not. – Dustan Levenstein Sep 10 '14 at 3:40
• @SDevalapurkar: How is $\mathbb{R} \times \mathbb{R}$ isomorphic to $\mathbb{R}$ as an additive group? – Doug Sep 10 '14 at 3:47
• @DanDouglas note that this is the direct sum, which consists of sequences such that all but finitely elements are equal to zero. That it, it is effectively the group of "terminating" sequences. – Omnomnomnom Sep 10 '14 at 3:52
• @DanDouglas in fact $\Bbb R\times\Bbb R\cong\Bbb R$ are isomorphic as vector spaces over $\Bbb Q$, not just as abelian groups. This is essentially because $\Bbb R$ has dimension $\frak c$ over $\Bbb Q$ and that $\frak c+c=c$. However we cannot write down such an isomorphism explicitly, because we invoke the axiom of choice to conclude it exists at all. – anon Sep 10 '14 at 3:56

For an example, consider $G=\mathbb Z[x]$ as an additive group. Then $G\times \mathbb Z \cong G\times\mathbb Z\times\mathbb Z$ but $\mathbb Z\not\cong \mathbb Z\times \mathbb Z$.
• By $\mathbb{Z}[x]$ do you mean single variable polynomials with coefficients in $\mathbb{Z}$? – Doug Sep 10 '14 at 3:59
• Am I correct that your map looks like, as an example, $(x + 3, 2, 5) \mapsto (x^2 + 3x + 2, 5)$? – Doug Sep 10 '14 at 4:05
• In fact, we could simply state $G \cong G \times \Bbb Z$ – Omnomnomnom Sep 10 '14 at 4:05