This is an assignment question: I just need know whether I am on the right track.....

Let $G_1,…,G_n$ be groups. Prove that the direct product $G_1×⋯×G_n$ is abelian if, and only if, each of $G_1,…,G_n$ is abelian. I know that this question have already been discussed before, I've got two thing that I need clarification on.... Is it wise for me to start this prove by assuming that $G_1,....,G_n$ (which is for each group) are Abelian, and work toward showing that the direct product of the group is abelian secondly, is there any second part of the prove that I haven't considered..

thank you... I'll really appreciate any help

  • $\begingroup$ You have to go both directions. You need to prove that the direct product being abelian implies that each $G_i$ is abelian too. Hint: consider the canonical projections out of the direct product. $\endgroup$ – Ian Coley Apr 3 '14 at 19:01
  • $\begingroup$ The first side is pretty easy. Assume for each $i$, $G_{i}$ is abelian. That means component $i$ in the direct product commutes on the operation. If each component commutes, the element in the product group commutes. For the converse, I would prove by contradiction. Suppose the direct product is abelian, but we have a group $G_{i}$ forming the product which is not abelian. You will get a contradiction that the product group is actually not abelian. $\endgroup$ – ml0105 Apr 3 '14 at 19:01
  • 2
    $\begingroup$ Hint of another sort: prove it for $n=2$ and then use induction. $\endgroup$ – drhab Apr 3 '14 at 19:08

For the "hard" direction:

$$A:=G_1\times\ldots\times G_n\;\;\text{abelian}\implies\;\;\text{any homomorphic image of $\,A\,$ is abelian}$$

and now just check the projection

$$\pi_i:A\to G_i\;\;,\;\;\;\pi_i(x_1,...,x_n):=x_i$$


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