0
$\begingroup$

Let finite group $G$ of order $2^n$ and $s$ an element of order $2$ in $G$. Is it always true that there exists a tower of subgroups $$(1) <G_1=(s)<G_2 < \ldots < G_n = G$$ and $[G_{i+1}\colon G_i]=2$ ?

If $s$ is in the center of $G$ that is clearly true.

If $G$ is a dihedral group $D_{2^n}$, then we have a sequence $$1< (s)=D_2<D_4< \ldots <D_{2^n}$$

I think the assertion is not true in general but I can't find a counterexample.

Edit: The assertion is true and much more, see the excellent answer of @xsnl: below.

$\endgroup$
  • 1
    $\begingroup$ It is true: a finite group is soluble iff it has a compositions series all of which factors are cyclic of prime order. Since a finite $\;p\,-$ group, $\;p\;$ a prime, is even nilpotent... $\endgroup$ – DonAntonio Apr 15 '18 at 17:27
  • $\begingroup$ @DonAntonio; Note that $(s)$ has to be the first term of the series. $\endgroup$ – Orest Bucicovschi Apr 15 '18 at 17:39
  • $\begingroup$ I didn't see that the first time, perhaps because of the non-standard notation...but it doesn't really matter: begin with $\;s=Z(G)\;$ and we're done. Only thing is that $\;|Z(G)|\;$ may not be of order two, of .course $\endgroup$ – DonAntonio Apr 15 '18 at 17:49
1
$\begingroup$

It's actually close to alternative characterization of nilpotent finite groups.

Theorem. Finite $G$ is nilpotent iff every subgroup of $G$ is subnormal.

Proof.

  1. Nilpotence implies subnormality.

Let $ H < G$. Consider upper central series of $G$: $1 = \zeta_0G \subset Z(G) = \zeta_1G \subset \dots \zeta_c G = G$ which is finite by nilpotence. As it is central series, $H\zeta_iG$ is normal in $H\zeta_{i+1}G$ so every subgroup is subnormal.

  1. Subnormality implies nilpotence.

Group with normal Sylows is nilpotent (as product of nilpotent groups). Consider non-normal Sylow $p$-subgroup of $G$; by standard Sylow package$^*$ every subgroup containing normalizer of Sylow is self-normalizing, therefore is not subnormal, contradiction.

$^*$: let $P$ be Sylow in $G$ and $N_G(P) \leq K \leq G$. If $a^{-1}Ka = K$, then $a^{-1}Pa = k^{-1}Pk$ for some $k \in K$ as $P$ is also Sylow in $K$ and Sylows are conjugate. But then $ak^{-1} \in N_G(P) \leq K$.



For $p$-group $P$ every $\Bbb Z/p \leq P$ is first term of composition series, because we've proven that every subgroup is subnormal and every subnormal chain in finite group refines to composition series. Composition series of $p$-group has $\Bbb Z/p$ factors, so your conjecture is true.



Sidenote on 2. For infinite groups, subnormality of all subgroups is weaker than nilpotence; but there are some results.

Theorem (Roseblade). If every subgroup of $G$ is subnormal by chain of length $C$ not depending on subgroup, then $G$ is nilpotent.

Theorem (Mohres). If every subgroup of $G$ is subnormal, then $G$ is solvable. (it's pretty hard to prove).

There's also an example by Heineken of a non-nilpotent group with every subgroup subnormal.

$\endgroup$
  • $\begingroup$ Thank you for your wonderful answer! $\endgroup$ – Orest Bucicovschi Apr 16 '18 at 0:57

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.