Is this problem stated correctly? Doesn't p have to be a prime for this to be true?

http://img832.imageshack.us/img832/9...10302at647.png

Similarly, doesn't p have to be prime?

http://img194.imageshack.us/img194/9...10302at647.png

- March 2nd 2011, 04:57 PMCropDusterGroups of order p^{e}m contains subgroups of order p^r for every integer r<=e.
Is this problem stated correctly? Doesn't p have to be a prime for this to be true?

http://img832.imageshack.us/img832/9...10302at647.png

Similarly, doesn't p have to be prime?

http://img194.imageshack.us/img194/9...10302at647.png - March 2nd 2011, 07:37 PMDrexel28
Yes should be prime. That said, this isn't a trivial trivial theorem per se. That said, unless you know the first Sylow theorem and the fact that -groups have a subgroup of every order dividing the group.

Quote: - March 2nd 2011, 07:47 PMtonio

6.4 - Yes, it must be p a prime in the first case:

, is a counterexample (p=6)

6.5 - The second question is false, too:

, is a counterexample (p=30) .

Even the condition is very weird, given the data of the question and,

of course, it should say "a propersubgroup, otherwise {1}__non-trivial__

makes the question itself trivial.

Or, of course, I'm missing something.

Tonio - March 4th 2011, 11:10 AMCropDuster
Ok, so I've been banging my head against the wall here for a while. I can't figure out how to proceed.

6.4

I know that there is an element in that has order . And I know that there is a Sylow -subgroup in , say such that . And I also know that the center of is not trivial. I also know that if is an element in , that is of order then is a positive power of . That is, if has order then , since Ord . But I just don't know how to proceed from here. Can anyone give me some hints? - March 4th 2011, 12:19 PMDrexel28
Let then by the first Sylow theorem there is a Sylow -subgroup, say . But, since and one has that (by the fact I said) that and thus has a subgroup of every and thus every .

Are you asking why a -group has a subgroup of every order dividing it? Try using the fact that the converse of Lagrange's theorem is true for abelian groups (or prove the result is true for abelian -groups...this is easy), the center of a -group is non-trivial and inducting on the power of ...this is one of many ways. Ask if you get stuck. - March 4th 2011, 01:31 PMCropDuster
I'm sorry, I think I left some ambiguity as to what assumption I can make. According to my text (Artin), a Sylow -subgroup of a group , where is a subgroup of that has order .

That said, the way the first Sylow theorem is stated in my book (and the way my class is using it) is as follows:

A finite group whose order is divisible by a prime contains a Sylow -subgroup.

So I feel like what you are suggesting depends on my*knowing*that there is a subgroup of order where . Based on the way we (my class/professor) are defining the first Sylow theorem, I don't think I can assume this. I realize that some books state the first Sylow theorem differently, and if I could apply that statement, then this problem would be proven just as you have suggested. - March 6th 2011, 11:08 AMDrexel28