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

1. ## 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?

Similarly, doesn't p have to be prime?

2. Originally Posted by CropDuster
Is this problem stated correctly? Doesn't p have to be a prime for this to be true?

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.

Similarly, doesn't p have to be prime?

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

3. Originally Posted by CropDuster
Is this problem stated correctly? Doesn't p have to be a prime for this to be true?

Similarly, doesn't p have to be prime?

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

4. 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?

5. 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.

6. 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.

7. 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.