We have already known that the inverse of Lagrange's Theorem is a right fact about for example abelian or nilpotent finite groups. How can I show that:

If $G$ be finite and supersolvable$^*$ and $n\mid|G|$, then $G$ has a subgroup of order $n$?

$^*$ A group is supersolvable if there exists normal subgroups $N_i$ with $$1=N_0\subseteq N_1\subseteq ...\subseteq N_r=G$$ where each factor $\frac{N_i}{N_{i-1}}$ is cyclic for $1\leq i\leq r$.

Thanks for any hint to start.

  • $\begingroup$ I like this question + $\endgroup$ – amWhy Feb 11 '13 at 4:25

See this article for a solution.

The basic idea is this. Suppose $G$ is supersolvable of order $$|G| = p_1^{a_1} \ldots p_{t-1}^{a_{t-1}} p_t^{a_t},$$ where $p_1 < p_2 < \ldots < p_t$ are primes. We can prove the claim by induction on $t$, the case $t = 1$ being true by Sylow's theorem. Because $G$ is solvable, its $p_t$-Sylow subgroup has a complement $H$. We can then use the inductive assumption on $H$ since subgroups of supersolvable groups are supersolvable. The claim follows by combining this with the fact that as a supersolvable group, $G$ has a normal subgroup of order $p_t^k$ for all $0 \leq k \leq a_t$.

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  • $\begingroup$ You may also see the note. $\endgroup$ – Shodharthi Nov 29 '13 at 11:21

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