# Thread: Subgroups of a group of order 2^m

1. ## Subgroups of a group of order 2^m

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

Can anyone help with this please?

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

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

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

3. 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 $G$ is a group of order $2^m$, then there exist subgroups

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

such that $|G_j| = 2^j$? 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 $p^n$ (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...)

4. Originally Posted by 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 $G$ is a group of order $2^m$, then there exist subgroups

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

such that $|G_j| = 2^j$? 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 $p^n$ (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 $2^n, \ n \geq 1,$ has a subgroup of order $2^{n-1}.$

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

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

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

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

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

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

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.

6. Originally Posted by Swlabr
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 $p^k \mid |G|,$ for some prime $p$ and integer $k \geq 0,$ then $G$ has a subgroup of order $p^k.$

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

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

8. Originally Posted by tonio
Perhaps some author includes this as part of Sylow theorems...?
Certainly Robinson doesn't...

9. 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.

10. Originally Posted by NonCommAlg
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.

11. well, i actually think that it should be a part of the theorem because it's a very natural extension of existence of Sylow subgroups.

12. Originally Posted by NonCommAlg
well, i actually think that it should be a part of the theorem because it's a very natural extension of existence of Sylow subgroups.

It may well be seen that way; nevertheless, most text books I know don't include it as part of Sylow theorems but as something that can more or less easily be proven without any reference to Sylow theorems, using only say the class equation and some induction.
Even in the case of the link you produced that part of what that author calls Sylow theorem is not proved within that theorem but apart, as lemma 4.11, without any reference to the actual Sylow theorem but only using precisely the fact that the center of finite p-groups is non-trivial! Thus, I'd say that including this fact within the Sylow theorem there is a pretty artificial thing.

It'd be interesting to know, me thinks, if Sylow included this result as part of his theorems or if it was known already.

Tonio

13. Thanks very much for this. Without worrying too much about exactly what the Sylow theorems say, I was looking at your more elementary proof and it's precisely the kind of argument I was after. One thing confuses me though: Cauchy's theorem doesn't require a group to be abelian, right? So why doesn't the proof you give for "case 1: G abelian" hold even when G isn't abelian?

On a related note, whilst I was searching for an answer to this problem I came across the comment, "The centre of any group of order $p^n$ (p prime) is not trivial. Then by Cauchy's theorem, we can pick an element in the centre of the group of order 2." So clearly there must be something to do with "commutativity" going on here, but I don't really understand what it is! Thanks.

14. Originally Posted by Boysilver
Thanks very much for this. Without worrying too much about exactly what the Sylow theorems say, I was looking at your more elementary proof and it's precisely the kind of argument I was after. One thing confuses me though: Cauchy's theorem doesn't require a group to be abelian, right? So why doesn't the proof you give for "case 1: G abelian" hold even when G isn't abelian?

On a related note, whilst I was searching for an answer to this problem I came across the comment, "The centre of any group of order $p^n$ (p prime) is not trivial. Then by Cauchy's theorem, we can pick an element in the centre of the group of order 2." So clearly there must be something to do with "commutativity" going on here, but I don't really understand what it is! Thanks.

On the contrary: thanx to the fact that the center of any finite p-group isn't trivial one doesn't need to worry about commutativity. For the induction step one only takes a non-trivial element of the center of the group and then looks at the subgroup generated by that element, which is normal for being central, and then passes to the factor group and etc.

Tonio