The problem is stated as this:

"Prove that if $$G:=G_1\times G_2 \times \dots\times G_s \cong H_1 \times H_2 \times \dots\times H_t=:H,$$ where each $G_i$ and $H_i$ is finite simple group, then $s=t$ and there exists $\sigma\in S_t$ such that $G_j \cong H_{\sigma(j)}$ for $j=1,\dots,t$."

This comes after first being introduced to concept of solvable groups and composition series. We are given the Jordan-Hölder Theorem as follows:

Any two composition series for a finite group have the same length. Further, there exists a one-to-one correspondence between composition factors of the two composition series under which corresponding composition factors are isomorphic.

As far as I understand, each $G_i$ being simple implies that the only composition series for $G$ is $$G \trianglelefteq e_G$$ where $e_G=(e_{G_1},\dots,e_{G_s})$, and similarly for $H$. Is this correct? If it is, I am unsure how to apply the Jordan-Hölder Theorem in this instance, since $G$ and $H$ are merely isomorphic.

Can anyone offer any hints or a solution? Thank you for help.


HINT: $$G_1 \le G_1 \times G_2 \le G_1 \times G_2 \times G_3 \le \dots \le G_1 \times \cdots \times G_n$$ is a composition series for $G$. You can see this noting that $G_1 \times \cdots \times G_{k-1} $ is an epimorphic image of $G_1 \times \cdots \times G_k$ simply by the projection, and the kernel is $1 \times \cdots \times 1 \times G_k \cong G_k$.

Now, do the same thing with the $H_i$s and use Jordan-Hölder.

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  • $\begingroup$ When you say $G_1 \le G_1\times G_2\times\dots$, do you mean $G_1\times { e_{G_2} }\times {e_{G_3}}\times \dots \times {e_{G_s}}\le G_1\times G_2\times {e_{G_3}}\times \dots \times {e_{G_s}}\le \dots $ ? Because $G_1$ is not a subgroup of $G_1 \times G_2$, is it? Also, can I apply Jordan-Hölder across two groups which are isomorphic? Thanks for the help.! $\endgroup$ – user279685 Oct 12 '15 at 22:24
  • $\begingroup$ Everybody thinks of $G$ as a subgroup of $G \times H$, although you are correct because the subgroup is actually $\{(g,1_H): g \in G \}$. Strangely enough, your question about isomorphic groups having isomorphic composition factors was asked very recently: math.stackexchange.com/questions/1477230 $\endgroup$ – Derek Holt Oct 13 '15 at 10:57

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