# Subgroups of a group of order 2^m

#### Boysilver

I'm trying to complete a standard proof concerning "ruler and compass constructibility" and it boils down to this result which I've seen stated in some books but haven't seen or been able to come up with a proof:

"If $$\displaystyle G$$ is a finite group of order $$\displaystyle 2^m$$ then there exist subgroups $$\displaystyle \{id\} \leq G_1 \leq G_2 \leq \ldots \leq G_n =G$$ such that $$\displaystyle |G_j| = 2^j$$."

Can anyone help with this please?

#### NonCommAlg

MHF Hall of Honor
I'm trying to complete a standard proof concerning "ruler and compass constructibility" and it boils down to this result which I've seen stated in some books but haven't seen or been able to come up with a proof:

"If $$\displaystyle G$$ is a finite group of order $$\displaystyle 2^m$$ then there exist subgroups $$\displaystyle \{id\} \leq G_1 \leq G_2 \leq \ldots \leq G_n =G$$ such that $$\displaystyle |G_j| = 2^j$$."

Can anyone help with this please?
this is actually true for any group of order $$\displaystyle p^m,$$ where $$\displaystyle p$$ is any prime number. the proof is by induction over $$\displaystyle m$$ and using this fact that every group of order $$\displaystyle p^m$$ has a subgroup of order $$\displaystyle p^{m-1}.$$ see

Sylow theorems! you can do even better than that: you can choose $$\displaystyle G_i$$ to be normal in $$\displaystyle G.$$

• Boysilver

#### Boysilver

I see, thanks. I've had a look at the Sylow theorems and they will solve my problem. From what I gather, my original question seems to be a particular case of a standard result. Just to check though, is there a quick and simple proof to show that when $$\displaystyle G$$ is a group of order $$\displaystyle 2^m$$, then there exist subgroups

$$\displaystyle \{id\} \leq G_1 \leq G_2 \leq \ldots \leq G_n$$

such that $$\displaystyle |G_j| = 2^j$$? Or really is the use of the Sylow theorems about the simplest argument we can give? Just because if there is indeed something simple that I could use instead, I'd much rather do it that way. I should have said: we can also use the fact that the centre of any group of order$$\displaystyle p^n$$ (p prime) is not trivial, and we can use Cauchy's theorem (if this helps, but I get the feeling they only help to prove the Sylow theorems anyway...)

#### NonCommAlg

MHF Hall of Honor
I see, thanks. I've had a look at the Sylow theorems and they will solve my problem. From what I gather, my original question seems to be a particular case of a standard result. Just to check though, is there a quick and simple proof to show that when $$\displaystyle G$$ is a group of order $$\displaystyle 2^m$$, then there exist subgroups

$$\displaystyle \{id\} \leq G_1 \leq G_2 \leq \ldots \leq G_n$$

such that $$\displaystyle |G_j| = 2^j$$? Or really is the use of the Sylow theorems about the simplest argument we can give? Just because if there is indeed something simple that I could use instead, I'd much rather do it that way. I should have said: we can also use the fact that the centre of any group of order$$\displaystyle p^n$$ (p prime) is not trivial, and we can use Cauchy's theorem (if this helps, but I get the feeling they only help to prove the Sylow theorems anyway...)
in math there's always another way! ok, so you want to do it without using Sylow theorems. we only need to prove that every group of order $$\displaystyle 2^n, \ n \geq 1,$$ has a subgroup of order $$\displaystyle 2^{n-1}.$$

we'll prove this by induction over $$\displaystyle n.$$ it's clear for $$\displaystyle n=1.$$ now suppose the claim is true for all groups of order $$\displaystyle 2^m$$ with $$\displaystyle 1 \leq m < n$$ and let $$\displaystyle G$$ be a group of order $$\displaystyle 2^n.$$ consider two cases:

case 1. $$\displaystyle G$$ is abelian: by Cauchy, $$\displaystyle G$$ has an element $$\displaystyle g$$ of order 2. let $$\displaystyle H=\langle x \rangle$$ and $$\displaystyle G_1=G/H.$$ then $$\displaystyle |G_1|=2^{n-1}$$ and so, by the induction hypothesis, $$\displaystyle G_1$$ has a subgroup of order $$\displaystyle 2^{n-2},$$

i.e. there exists a subgroup $$\displaystyle N \supset H$$ of $$\displaystyle G$$ such that $$\displaystyle |N/H|=2^{n-2}$$ and thus $$\displaystyle |N|=2^{n-2}|H|=2^{n-1}.$$

case 2. $$\displaystyle G$$ is non-abelian: so $$\displaystyle |Z(G)|=2^k,$$ for some integer $$\displaystyle 1 \leq k < n.$$ let $$\displaystyle G_1=G/Z(G).$$ then $$\displaystyle 2 \leq |G_1|=2^{n-k} < 2^n=|G|.$$ thus, by the induction hypothesis, $$\displaystyle G_1$$ has a subgroup of

order $$\displaystyle 2^{n-k-1},$$ i.e. there exists a subgroup $$\displaystyle N \supset Z(G)$$ of $$\displaystyle G$$ such that $$\displaystyle |N/Z(G)|=2^{n-k-1}$$ and hence $$\displaystyle |N|=2^{n-k-1}|Z(G)|=2^{n-1}.$$

#### Swlabr

I'm trying to complete a standard proof concerning "ruler and compass constructibility" and it boils down to this result which I've seen stated in some books but haven't seen or been able to come up with a proof:

"If $$\displaystyle G$$ is a finite group of order $$\displaystyle 2^m$$ then there exist subgroups $$\displaystyle \{id\} \leq G_1 \leq G_2 \leq \ldots \leq G_n =G$$ such that $$\displaystyle |G_j| = 2^j$$."

Can anyone help with this please?
I am a tad confused by the Sylow's theorem discussion - Sylow's Theorems don't work in p-groups...I mean, you are trying to find the largest p-subgroup, so it is just the whole group!

Also, the result is proven in the literature, for example on page 139 of Robinson, A Course in the Theory of Groups.

#### NonCommAlg

MHF Hall of Honor
I am a tad confused by the Sylow's theorem discussion - Sylow's Theorems don't work in p-groups...I mean, you are trying to find the largest p-subgroup, so it is just the whole group!

Also, the result is proven in the literature, for example on page 139 of Robinson, A Course in the Theory of Groups.
one part of Sylow theorems says that if $$\displaystyle p^k \mid |G|,$$ for some prime $$\displaystyle p$$ and integer $$\displaystyle k \geq 0,$$ then $$\displaystyle G$$ has a subgroup of order $$\displaystyle p^k.$$

#### tonio

one part of Sylow theorems says that if $$\displaystyle p^k \mid |G|,$$ for some prime $$\displaystyle p$$ and integer $$\displaystyle k \geq 0,$$ then $$\displaystyle G$$ has a subgroup of order $$\displaystyle p^k.$$

I may be wrong but I don't remember any Sylow theorem stating that. Of course, if there's a Sylow subgroup then this follows from what the OP is trying to prove.
Perhaps some author includes this as part of Sylow theorems...?

Tonio

#### Swlabr

Perhaps some author includes this as part of Sylow theorems...?
Certainly Robinson doesn't...

#### NonCommAlg

MHF Hall of Honor
well, Sylow theorems have been stated in different ways. i don't have access to any group theory textbook right now and so i googled the Sylow theorems and the first noteable thing was a

lecture note of some professor in the university of Glasgow and this guy agrees with me. see theorem 4.10. anyway, the proof is the same as the proof i gave for p = 2.

#### Swlabr

well, Sylow theorems have been stated in different ways. i don't have access to any group theory textbook right now and so i googled the Sylow theorems and the first noteable thing was a

lecture note of some professor in the university of Glasgow and this guy agrees with me. see theorem 4.10. anyway, the proof is the same as the proof i gave for p = 2.
I would be inclined to disagree with that statement of the Theorem. I have never seen that theorem included in Sylows Theorems, and I suspect it is just put in there because it is both a nice result and relevant. I doubt it would be in a `classical' statement of the theorem. Also, it stands alone in the theorems - it is not used in any other part of the theorem, and nor is any other part used to prove it.

I suppose it is just my own opinion though.