# Proving that Direct product $G_1 \times … \times G_n$ is abelian if and only if each of $G_1, …,G_n$ is abelian

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

how can we prove that the direct product $G_1 \times .... \times G_n$ is abelian if and only if each of $G_1,...,G_n$ is abelian.

Please if someone can help and guide with this question...

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Keep in mind that two elements of $G_1 \times .... \times G_n$ are equal if and only if all of their components are equal, and that the operation on $G_1 \times .... \times G_n$ is made componentwise.

So $(a_1, \dots, a_n) \cdot (b_1, \dots, b_n) = (b_1, \dots, b_n) \cdot (a_1, \dots, a_n)$ if and only if...

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+1 for making another taste. – Babak S. Feb 1 '13 at 10:15
Thankx a lot for your guidance and help.. – Denish Sen Feb 1 '13 at 22:59
@DenishSen, you're welcome! – Andreas Caranti Feb 1 '13 at 23:48
Do we have to complete our proof by saying that the inverse and unit elements exists ...I mean if we write if and only if the element (e, e) is a unit element since for any (a,b) Ɛ G1,…….Gn holds that (a,b).(e,e) = (a.e, b.e) = (a, b) and similarly (e,e)(a,b) = (a,b). The inverse of typical element (a, b) Ɛ G1 x …….Gn is (a-1, b-1) since by definition, (a,b) . (a-1, b-1) = (a.a-1, b.b-1) = (e, e) and similarly , (a-1, b-1). (a,b) = (e, – Denish Sen Feb 7 '13 at 10:40
@DenishSen, it depends on the assignment. From your formulation, it appears that the construction of the direct product is already assumed. In any case, the proofs you mention are indeed done componentwise. – Andreas Caranti Feb 7 '13 at 10:55

The forward direction should be clear. For the reverse direction, suppose there is some $G_i$ which is not abelian, meaning that there are $a,b\in G_i$ such that $ab\neq ba.$

Now consider $xy$ and $yx$ where $x= (1,1,\cdots, a, \cdots, 1), y= (1,1,\cdots, b, \cdots, 1) \in G_1 \times G_2 \cdots \times G_n$. What does this say about the abelian-ness of the direct product?

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Why so indirect? You can prove it directly. – Martin Brandenburg Feb 1 '13 at 11:16
@MartinBrandenburg Ahh that's true. For some reason this was how I thought of it first, but now that you point it out, the direct proof is more natural. – Ragib Zaman Feb 1 '13 at 11:58
Thank you Sir for the help... – Denish Sen Feb 2 '13 at 9:12