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

- Mar 2nd 2011, 03:57 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 - Mar 2nd 2011, 06:37 PMDrexel28
Yes $\displaystyle p$ 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 $\displaystyle p$-groups have a subgroup of every order dividing the group.

Quote: - Mar 2nd 2011, 06:47 PMtonio

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

$\displaystyle A_4\,,\,\,|A_4|=6\cdot 2$ , is a counterexample (p=6)

6.5 - The second question is false, too:

$\displaystyle A_5\,,\,\,|A_5|=30\cdot 2$ , is a counterexample (p=30) .

Even the condition $\displaystyle |G|>p$ 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 - Mar 4th 2011, 10:10 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 $\displaystyle G$ that has order $\displaystyle p$. And I know that there is a Sylow $\displaystyle p$-subgroup in $\displaystyle G$, say $\displaystyle H\subseteq G$ such that $\displaystyle |H|=p^e$. And I also know that the center of $\displaystyle H$ is not trivial. I also know that if $\displaystyle x$ is an element in $\displaystyle G$, that is of order $\displaystyle p$ then $\displaystyle x$ is a positive power of $\displaystyle p$. That is, if $\displaystyle x^k$ has order $\displaystyle p$ then $\displaystyle k=p^{e-1}$, since Ord$\displaystyle (x^k)=\frac{|H|}{\gcd\bigl[k,|H|\bigr]}=p$. But I just don't know how to proceed from here. Can anyone give me some hints? - Mar 4th 2011, 11:19 AMDrexel28
Let $\displaystyle p^k\| p^e m$ then by the first Sylow theorem there is a Sylow $\displaystyle p$-subgroup, say $\displaystyle H$. But, since $\displaystyle |H|=p^k$ and $\displaystyle e\leqslant k$ one has that (by the fact I said) that $\displaystyle H$ and thus $\displaystyle G$ has a subgroup of every $\displaystyle \ell\leqslant k$ and thus every $\displaystyle \ell\leqslant e$.

Are you asking why a $\displaystyle p$-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 $\displaystyle p$-groups...this is easy), the center of a $\displaystyle p$-group is non-trivial and inducting on the power of $\displaystyle p$...this is one of many ways. Ask if you get stuck. - Mar 4th 2011, 12:31 PMCropDuster
I'm sorry, I think I left some ambiguity as to what assumption I can make. According to my text (Artin), a Sylow $\displaystyle p$-subgroup of a group $\displaystyle G$, where $\displaystyle |G|=p^{e}m$ is a subgroup of $\displaystyle G$ that has order $\displaystyle p^e$.

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 $\displaystyle p$ contains a Sylow $\displaystyle p$-subgroup.

So I feel like what you are suggesting depends on my*knowing*that there is a subgroup of order $\displaystyle p^k$ where $\displaystyle k\leq e$. 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. - Mar 6th 2011, 10:08 AMDrexel28