4
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ I like this question + $\endgroup$ – amWhy Feb 11 '13 at 4:25
2
$\begingroup$

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 $0 \leq k \leq a_t$.

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.