# Tower of subgroups for a finite $2$-group

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.

• 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... – DonAntonio Apr 15 '18 at 17:27
• @DonAntonio; Note that $(s)$ has to be the first term of the series. – Orest Bucicovschi Apr 15 '18 at 17:39
• 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 – DonAntonio Apr 15 '18 at 17:49

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.

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