# Thread: nilpotent groups

1. ## nilpotent groups

I need to show subgroups of nilpotents groups are nilpotent.

I can get started but cannot finish.

Suppose G is a nilpotent group. Let H be a proper subgroup of G. Let K be a proper subgroup of H. I must show K is a proper subgroup of $\displaystyle N_H(K)$.

I assume $\displaystyle N_H(K)=K$. I know I must get a contradiction here but I can't figure it out.

Any hints?

2. Originally Posted by ziggychick
I need to show subgroups of nilpotents groups are nilpotent.

I can get started but cannot finish.

Suppose G is a nilpotent group. Let H be a proper subgroup of G. Let K be a proper subgroup of H. I must show K is a proper subgroup of $\displaystyle N_H(K)$.

I assume $\displaystyle N_H(K)=K$. I know I must get a contradiction here but I can't figure it out.

Any hints?
what you're trying to prove doesn't prove that H is nilpotent unless G is finite. is G a finite group?

3. in this case, yes G is finite. didn't realize that was an issue.. :-/

4. Originally Posted by ziggychick

in this case, yes G is finite. didn't realize that was an issue.. :-/
yes, the "normalizer condition" that you're trying to prove is equivalent to "nilpotent" only for finite groups. but the fact is that a subgroup of any nilpotent group is nilpotent and we prove

results about nilpotent groups using central series not the normalizer condition!

5. do you only know the definition of "finite nilpotent groups" or you're also familiar with lower and upper central series, which give the definition of "nilpotent" for general (finite or infinite) groups?

6. my text is focused on finite nilpotent groups but it mentions that most of the results hold for infinite ones. i'm familiar with upper and lower central series, but we haven't really done any work with them.

7. Originally Posted by ziggychick

my text is focused on finite nilpotent groups but it mentions that most of the results hold for infinite ones. i'm familiar with upper and lower central series, but we haven't really done any work with them.
ok! can we use this theorem that a finite group G is nilpotent iff every Sylow subgroup of G is normal?

8. yes we can definitely use that.
i tried to go that route as well, but maybe i didn't think hard enough about it.

Suppose H is a proper subgroup of finite nilpotent group G. Let Q be a Sylow p-subgroup of H. Then we get Q is a subgroup of the Sylow p-subgroup P of G. We know P is normal in G, and is unique.

If $\displaystyle |Q|=p^k$ can P have another subgroup of the same order?

9. Originally Posted by ziggychick
yes we can definitely use that.
i tried to go that route as well, but maybe i didn't think hard enough about it.

Suppose H is a proper subgroup of finite nilpotent group G. Let Q be a Sylow p-subgroup of H. Then we get Q is a subgroup of the Sylow p-subgroup P of G. We know P is normal in G, and is unique.

If $\displaystyle |Q|=p^k$ can P have another subgroup of the same order?
well, $\displaystyle P \cap H$ is a p subgroup of $\displaystyle H$ and it contains $\displaystyle Q.$ thus $\displaystyle P \cap H=Q,$ because $\displaystyle Q$ is a Sylow p subgroup of $\displaystyle H.$ can you finish the proof now?

10. yes, i believe i can. thanks so much!

11. Originally Posted by NonCommAlg

the "normalizer condition" that you're trying to prove is equivalent to "nilpotent" only for finite groups.
just to correct myself (for the sake of completeness): the "normalizer condition" is equivalent to "nilpotent" in "finitely generated" groups and not only in finite groups.