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This is another proof question I am asking about, can someone give me tips on how to answer these questions?

My question says:

"Let $H,G$ be arbitrary groups. Prove that $H \times G$ is commutative (abelian) if and only if both groups $H, G$ are commutative"

I don't get how to prove that $H $ and $G$ are commutative in general. Seeing as they are arbitrary groups, they can be anything right? How do I go about trying to prove this?

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up vote 6 down vote accepted

In general, be aware of and state what you know.

In one direction you know that $H \times G$ is commutative. What does it mean? It means that for all $(h_1, g_1), (h_2, g_2) \in H \times G$ you have $(h_1, g_1) + (h_2, g_2) = (h_1 + h_2, g_1 + g_2) = (h_2, g_2) + (h_1, g_1)$.

Next be aware of and state what you want. You want to show that for all $h_1, h_2 \in H, g_1, g_2 \in G$ you have $h_1 + h_2 = h_2 + h_1$ and $g_1 + g_2 = g_2 + g_1$.

To finish, use what you know to show what you want.

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  • The direct product $H\times K$ is defined so that it contains subgroups (isomorphic to) $H$ and $K$, and that all elements of the first subgroup commute with all elements of the second one, and the union of these subgroups generates the whole group.

  • For any set of elements that all commute among each other, all products of such elements also commute among each other.

  • Any subgroup of an Abelian group is Abelian

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