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Let $G$ be a finite group. If $G= G_1\times G_2 \cdots G_t$ and $G = H_1 \times H_2 \cdots H_t$ with each $G_j,H_j$ indecomposable , then $s=t$ and after reindexing $G_i \cong H_i$ for every $i$ and for each $r \le t$

$G = G_1 \times G_2 \cdots G_r \times H_{r+1}\times \cdots H_t$

Notice that the uniqueness statement is stronger than simply saying that the indecomposable factors are determined upto isomorphism.

Question 1 : I am not getting the meaning of bold text.

Question 2 : Why condition that each factor has to be $G_i$ indecomposable?

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  • $\begingroup$ 1. I think it means that you do not only have $\{G_1,\dots,G_t\}=\{H_1,\dots,H_t\}$ (pretty much like you have $\{1,1,2\}=\{1,2,2\}$) but you really have the same elements up to reindexing the sequences (i.e. the same indecomposable factors up to reindexation). $\endgroup$ – Clément Guérin Mar 16 '18 at 7:56
  • $\begingroup$ 2. Consider $G=\mathbb{Z}/2\times\mathbb{Z}/2\times \mathbb{Z}/4$, you could take $G_1=\mathbb{Z}/2\times\mathbb{Z}/2$, $G_2=\mathbb{Z}/4$, $H_1=\mathbb{Z}/2$ and $H_2=\mathbb{Z}/2\times\mathbb{Z}/4$. Whence indecomposable is an important hypothesis. $\endgroup$ – Clément Guérin Mar 16 '18 at 7:57

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