If we let $G_1,...,G_n$ be groups,

When proving that the direct product $G_1 \times .... \times G_n$ is abelian if and only if each of $G_1,...,G_n$ is abelian, can someone please help me Im concerned about whether it should also prove that it holds for the conditions of a group to be abelian(inverse, unit element ...) or just prove straightforward that left hand side is true iff right hand side is?

Can someone please help clarify this. Thanks


marked as duplicate by Seirios, Gerry Myerson, user26857, Alexander Gruber, Hagen von Eitzen Feb 9 '13 at 9:19

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There are two parts that you need to prove, as follows (in either order).

First, assume that the direct product $G_{1}\times\cdots\times G_{n}$ is abelian, and then show that each of $G_{1},\ldots, G_{n}$ must also be abelian. (Hint: Find a homomorphism from the direct product onto an arbitrary $G_{i}$ and use that to help with the proof.)

Second, assume that all of $G_{1},\ldots, G_{n}$ are abelian, and show that the direct product $G_{1}\times\cdots\times G_{n}$ must be abelian. (Hint: Just use the definitions. What does a typical element of $G_{1}\times\cdots\times G_{n}$ look like?)

  • 2
    $\begingroup$ for the first part also just working with the definitions work. This is just to clarify that there is nothing fancy needed for either parts. $\endgroup$ – Ittay Weiss Feb 9 '13 at 8:23
  • $\begingroup$ Agreed. I just have this penchant for trying to get students to think with homomorphisms. :-) $\endgroup$ – James Feb 9 '13 at 8:45
  • $\begingroup$ Thank you both for your feedback. $\endgroup$ – Faye Feb 9 '13 at 8:57
  • $\begingroup$ Great, so its just using the definitions for the direct products of $n$ groups and not prove the axioms for a group to be abelian. Right. Cheers. $\endgroup$ – Faye Feb 9 '13 at 9:01

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