Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

Let $G_1,\dots,G_n$ be groups. Prove that the direct product $G_1\times\cdots\times G_n$ is abelian if, and only if, each of $G_1,\dots,G_n$ is abelian. To prove that the direct product is abelian is straightforward but what I don't understand is the converse.

share|cite|improve this question
up vote 4 down vote accepted

The main point in this problem is that two elements of $$G_1 \times \cdots \times G_n$$ are equal iff their components are equal. You can use this fact for showing both directions. In fact since the operation on $$G_1 \times \cdots \times G_n$$ is made component-wise. So $$(a_1, \dots, a_n) \cdot (b_1, \dots, b_n) = (b_1, \dots, b_n) \cdot (a_1, \dots, a_n)$$ then $$(a_1b_1, \dots, a_nb_n) = (b_1a_1, \dots, b_na_n)$$ and so $$\forall ~i, ~a_ib_i=b_ia_i$$ and vice versa.

share|cite|improve this answer
    
Nice! Straightforward...+1 – amWhy Feb 8 '13 at 5:50

HINT: $G_1 \times \cdots \times G_n \rightarrow G_i: (g_1,\ldots,g_n) \mapsto g_i$ is a homomorphism and the homomorphic image of an abelian group is abelian.

share|cite|improve this answer
    
Yes, although this won't help OP if the course hasn't done homomorphisms yet. – Gerry Myerson Feb 8 '13 at 4:31
    
@GerryMyerson Well, then he should say so shouldn't he :) – sxd Feb 8 '13 at 4:33
    
@ Dimitri Surinx yes we've learn homomorphism. thank you – Bulou Duikoro Feb 9 '13 at 0:33

Your Answer

 
discard

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.