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For a group $G$ the length $l(G)$ is defined as the supremum of the lengths of all composition series of $G$. If $N$ is a normal subgroup of $G$, then $l(G)=l(N)+l(G/N)$. This, I believe, is a consequence of the Jordan-Hölder theorem.

For a subgroup $H$ of $G$ that is not normal in $G$, maybe even self-normalizing, is it still true that $l(H)\leq l(G)$?

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up vote 10 down vote accepted

No. For instance, for sufficiently large $n$ you can have arbitrarily long composition series in subgroups of the symmetric group $S_n$, which has composition length $2$ for $n\ge5$. The subgroup generated by the disjoint $2$-cycles $(12), (34), (56), \dotsc, ((2k-1)2k)$ is isomorphic to $\mathbb Z_2^k$, whose composition length is $k$.

More generally, every finite group $G$ can be regarded as a subgroup of the symmetric group $S_{|G|}$, so there wouldn't be any composition series of length greater than $2$. As a further example, $S_4$ has composition length $4$ and can be regarded as a subgroup of $S_5$, which has composition length $2$.

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Thank you very much. – Stefan Jul 16 '11 at 14:22

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