Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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...

share|cite|improve this question

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...

share|cite|improve this answer
+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?

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.