"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?

"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

"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?

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

\(\displaystyle \{id\} \leq G_1 \leq G_2 \leq \ldots \leq G_n \)

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 \(\displaystyle 2^{n-1}.\)

\(\displaystyle \{id\} \leq G_1 \leq G_2 \leq \ldots \leq G_n \)

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...)

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:

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

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 \(\displaystyle |N|=2^{n-k-1}|Z(G)|=2^{n-1}.\)

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!

"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?

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 \(\displaystyle p^k.\)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 \(\displaystyle p^k.\)

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

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

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.

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 suppose it is just my own opinion though.