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$\begingroup$

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.

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    Consider $\mathbb{R}^2$.2012-09-09
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    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.2012-09-09
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    One example where the groups are distinct is $\mathbb{Z}_2 \times \mathbb{Z}_4$.2012-09-09
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    @ChrisEagle: May I ask what do you mean by group $\mathbb R$? Do you mean $\mathbb R^{+}$ or $\mathbb R^{\times}$? Thank you.2012-09-10
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    Consider whichever you like, they both work.2012-09-10
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    But ${\mathbb R}$ isn't a group under multiplication!2012-09-10
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    @DerekHolt: Dear Prof. I meant $\mathbb R^*$. But your example is so understandable for me. Of course you left for me finding the conditions under which they are unique. Thank you very much for the time.2012-09-10
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