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Which finite groups have uniqueness for the ordered sequence of composition factors (up to isomorphism)?

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If you look at the proof of uniqueness of the set of composition factors, you will easily find out when it gives uniqueness of ordered sequences. Basically the "only" example when non-uniqueness occurs is the direct product of two simple groups. –  Mark Sapir Aug 17 '12 at 15:56
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@Mark: what is a good source for this proof? –  Igor Rivin Aug 17 '12 at 18:09
    
@Igor: Any book on group theory, that covers finite groups. For example, M. Hall "Group theory", Chapter 8. –  Mark Sapir Aug 17 '12 at 19:18
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1 Answer

up vote 12 down vote accepted

Here is a characterization. A group $G$ has two different composition series if and only if it has a factor $H/K$ which is a direct product of two non-isomorphic simple subgroups, where $H$ is a subnormal subgroup of $G$, $K$ is a normal subgroup of $H$. Indeed, if such $H/K$ exists, then clearly there are two different composition series. Conversely, suppose that there are two different composition series $A_0=1 < A_1 < A_2 < ... < A_n = G$ and $B_0=1 < B_1 < B_2 < ... < B_n=G$. Let $j$ be the last index with $A_{j-1}\ne B_{j-1}$ (non-isomorphic), $n\ge j\ge 1$. Let $H=A_j=B_j$. Note that $A_{j-1}$ and $B_{j-1}$ are normal in $H$. Hence $A_{j-1}B_{j-1}$ is normal in $H$. Since it is bigger than $A_{j-1}$, we have $H=A_{j-1}B_{j-1}$. Hence $K=A_{j-1}\cap B_{j-1}$ is normal in $H$ and $A_{i-1}/K$ is isomorphic to $H/B_{j-1}$ hence simple. Similarly $B_{j-1}/K$ is simple. Now $H/K$ has two different normal simple subgroups $A_{j-1}/K$ and $B_{j-1}/K$ with trivial intersection, so $A_{j-1}/K$ and $B_{j-1}/K$ commute and form a direct product. Therefore $H/K$ is isomorphic to the direct product of two non-isomorphic simple groups $A_{j-1}/K$ and $B_{j-1}/K$.

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It seems to me that this proof should be turned upside down. I don't think that $A_{j+1}/K$ and $B_{j+1}/K$ need be normal in $H/K$ in the given set-up. For example, say $G$ is dihedral of order eight and $A_1,B_1$ are generated by two different reflections that do not commute. Then $K=1$ and $H=G$. If, on the other hand, you find the largest $j$ such that $A_j \neq B_j$ and consider $A_{j+1}/(A_j \cap B_j)$, you should be in business. –  John Shareshian Aug 18 '12 at 1:55
    
@John: you are right. I will change the answer. –  Mark Sapir Aug 18 '12 at 2:30
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You mean: "which is a direct product of two non-isomorphic simple subgroups". Or I misunderstood the question (which is vaguely stated)... I understand that if $S$ is simple, $S^2$ has uniqueness of the ordered composition series up to isomorphism (of the simple factors), namely $(S,S)$. –  Yves Cornulier Aug 18 '12 at 6:33
    
Yes, non-isomorphic. I have added that. –  Mark Sapir Aug 18 '12 at 8:14
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