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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.

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If this is homework you should tag it as such. Also, you might want to include the steps you take and where you got stuck to because from what you had written it is not clear where you are stuck. –  Ittay Weiss Feb 8 '13 at 3:50
Did I miss understand your question? Which direction do you want help with? –  sxd Feb 8 '13 at 4:21
@ Ittay Weiss proving that the direct product is abelian that i understood but to prove that in the reverse way that i don't understand –  Bulou Duikoro Feb 9 '13 at 0:40

2 Answers 2

up vote 3 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.

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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.

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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

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