# Groups of order p^{e}m contains subgroups of order p^r for every integer r<=e.

• March 2nd 2011, 03:57 PM
CropDuster
Groups 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, 06:37 PM
Drexel28
Quote:

Originally Posted by CropDuster
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

Yes $p$ 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 $p$-groups have a subgroup of every order dividing the group.

Quote:

Similarly, doesn't p have to be prime?

http://img194.imageshack.us/img194/9...10302at647.png
Yes, similarly you should assume that $p$ is prime here too. Also, not a trivial exercise.
• March 2nd 2011, 06:47 PM
tonio
Quote:

Originally Posted by CropDuster
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

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

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

6.5 - The second question is false, too:

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

Even the condition $|G|>p$ is very weird, given the data of the question and,

of course, it should say "a proper non-trivial subgroup, otherwise {1}

makes the question itself trivial.

Or, of course, I'm missing something.

Tonio
• March 4th 2011, 10:10 AM
CropDuster
Quote:

Originally Posted by Drexel28
Yes $p$ 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 $p$-groups have a subgroup of every order dividing the group.

Yes, similarly you should assume that $p$ is prime here too. Also, not a trivial exercise.

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 $G$ that has order $p$. And I know that there is a Sylow $p$-subgroup in $G$, say $H\subseteq G$ such that $|H|=p^e$. And I also know that the center of $H$ is not trivial. I also know that if $x$ is an element in $G$, that is of order $p$ then $x$ is a positive power of $p$. That is, if $x^k$ has order $p$ then $k=p^{e-1}$, since Ord $(x^k)=\frac{|H|}{\gcd\bigl[k,|H|\bigr]}=p$. But I just don't know how to proceed from here. Can anyone give me some hints?
• March 4th 2011, 11:19 AM
Drexel28
Quote:

Originally Posted by CropDuster
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 $G$ that has order $p$. And I know that there is a Sylow $p$-subgroup in $G$, say $H\subseteq G$ such that $|H|=p^e$. And I also know that the center of $H$ is not trivial. I also know that if $x$ is an element in $G$, that is of order $p$ then $x$ is a positive power of $p$. That is, if $x^k$ has order $p$ then $k=p^{e-1}$, since Ord $(x^k)=\frac{|H|}{\gcd\bigl[k,|H|\bigr]}=p$. But I just don't know how to proceed from here. Can anyone give me some hints?

Let $p^k\| p^e m$ then by the first Sylow theorem there is a Sylow $p$-subgroup, say $H$. But, since $|H|=p^k$ and $e\leqslant k$ one has that (by the fact I said) that $H$ and thus $G$ has a subgroup of every $\ell\leqslant k$ and thus every $\ell\leqslant e$.

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

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

So I feel like what you are suggesting depends on my knowing that there is a subgroup of order $p^k$ where $k\leq e$. 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.
• March 6th 2011, 10:08 AM
Drexel28
Quote:

Originally Posted by CropDuster
I'm sorry, I think I left some ambiguity as to what assumption I can make. According to my text (Artin), a Sylow $p$-subgroup of a group $G$, where $|G|=p^{e}m$ is a subgroup of $G$ that has order $p^e$.

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

So I feel like what you are suggesting depends on my knowing that there is a subgroup of order $p^k$ where $k\leq e$. 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.

If you look at my suggestion it's still applicable. What I suggest is using the fact (Sylow's first theorem mine and yours) that a finite group has a subgroup of the maximal power of any prime dividing the order.