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We know that:

Theorem: if $G=\prod_{i=1}^n G_i$ be the direct product of groups $G_1,G_2,...,G_n$ then there exists normal subgroups $\bar{G_i}\cong G_i( i=1...n)$ of $G$ such that $$G=\bar{G_1}\bar{G_2}...\bar{G_n}$$ and $$\forall i, 1\leq i\leq n; \bar{G_i}\cap(\bar{G_1}...\bar{G}_{i-1}\bar{G}_{i+1}...\bar{G_n})=\{1\}$$

Maybe my question is so simple but: Could $\bar{G_i}$ be uniquely chosen, however, we don’t have this , stated in above theorem? Thanks.

share|cite|improve this question
Consider $\mathbb{R}^2$. – Chris Eagle Sep 9 '12 at 9:07
Or consider the Klein 4-group. You should always think about some small examples before making guesses. But it would be a good exercise to find some conditions under which they are unique - for example if they are finite groups of mutually coprime order. – Derek Holt Sep 9 '12 at 9:53
One example where the groups are distinct is $\mathbb{Z}_2 \times \mathbb{Z}_4$. – Mikko Korhonen Sep 9 '12 at 13:22
@ChrisEagle: May I ask what do you mean by group $\mathbb R$? Do you mean $\mathbb R^{+}$ or $\mathbb R^{\times}$? Thank you. – Babak S. Sep 10 '12 at 6:36
Consider whichever you like, they both work. – Chris Eagle Sep 10 '12 at 9:15

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