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If $G$ and $H$ are groups, $G\cong H$, and $$G=N_{0}\geq N_1\geq \dots \geq N_s ={e_G}$$ and $$H=M_{0}\geq M_1\geq \dots \geq M_t ={e_H}$$ are composition series for $G$ and $H$, what can be said about the relationship between the $N_i$s and the $M_i$s?

By Jordan-Holder, we know that any two composition series for a particular group have the same length, and that there is a 1-1 correspondence between factors of any two composition series under which corresponding factors are isomorphic.

Does the isomorphism $G\cong H$ allow us to conclude that $s=t$ in the above example, and that each $N_i$ is isomorphic to some $M_j$ and vice-versa? If so, can you explain explicitly how it follows from Jordan-Holder or otherwise?

Thanks for the help, I am new to this concept.

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  • $\begingroup$ Yes, it follows from Jordan-Holder. Show that you can use an isomorphism between two groups $G$ and $H$ to write down a bijection between their composition series that respects factors. $\endgroup$ – Qiaochu Yuan Oct 12 '15 at 23:41
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Because of isomorphism in $G$ and $H$, your question can be reduced to simple one: if we consider two composition series of same group $G$, what can be said about them? The answer, you know well, is that the composition factors are isomorphic.

Hence now instead of saying $G$ and $H$ are same, if we consider them as isomorphic, then still same answer comes in terms of isomorphism: composition factors of $G$ and $H$ are isomorphic up to permutation.

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