# Unique Sylow p-Subgroups

• Dec 3rd 2011, 12:28 PM
olgashukina
Unique Sylow p-Subgroups
If p is an odd prime, prove the following.
a) If G is a group of order $\displaystyle (p-1)p^2$, then G has a unique Sylow p-subgroup.
b) There are at least four groups of order $\displaystyle (p-1)p^2$which are pairwise nonisomorphic.

I know little about Sylow's subgroups.

$\displaystyle |G|=(p-1)p^2$

$\displaystyle n_i\equiv 1 \ (\text{mod} \ i)$
• Dec 3rd 2011, 02:53 PM
Drexel28
Re: Unique Sylow p-Subgroups
Quote:

Originally Posted by olgashukina
If p is an odd prime, prove the following.
a) If G is a group of order $\displaystyle (p-1)p^2$, then G has a unique Sylow p-subgroup.

You know that the number of Sylow $\displaystyle p$-subgroups of $\displaystyle G$ divides $\displaystyle p-1$ and is equivalent to $\displaystyle 1$ modulo $\displaystyle p$. Now, tell me, how many numbers strictly less than any given $\displaystyle n$ are equivalent to $\displaystyle n$ modulo $\displaystyle 1$?

Quote:

b) There are at least four groups of order $\displaystyle (p-1)p^2$which are pairwise nonisomorphic.

Ok, once you take care of the obvious abelian ones, the important observation is that $\displaystyle p-1\mid |\text{Aut}(\mathbb{Z}_{p^2})|=\varphi(p^2)=p(p-1)$ and so there exists non-trivial ____ products.
• Dec 3rd 2011, 04:13 PM
Deveno
Re: Unique Sylow p-Subgroups
it might be helpful to note that $\displaystyle \text{Aut}(\mathbb{Z}_{p^2}) \cong U(\mathbb{Z}_{p^2}) = (\mathbb{Z}_{p^2})^{\times}$

and here i'm a bit confused. it seems to me, that this only gives 3 subgroups of order $\displaystyle (p-1)p^2$ for the case p = 3.

so i believe that you also must investigate where you have a homomorphism into

$\displaystyle \text{Aut}(\mathbb{Z}_p \times \mathbb{Z}_p)$ as well.
• Dec 3rd 2011, 05:15 PM
Drexel28
Re: Unique Sylow p-Subgroups
Quote:

Originally Posted by Deveno
it might be helpful to note that $\displaystyle \text{Aut}(\mathbb{Z}_{p^2}) \cong U(\mathbb{Z}_{p^2}) = (\mathbb{Z}_{p^2})^{\times}$

and here i'm a bit confused. it seems to me, that this only gives 3 subgroups of order $\displaystyle (p-1)p^2$ for the case p = 3.

so i believe that you also must investigate where you have a homomorphism into

$\displaystyle \text{Aut}(\mathbb{Z}_p \times \mathbb{Z}_p)$ as well.

Right, $\displaystyle \text{Aut}(\mathbb{Z}_p^2)\cong \text{GL}_2(\mathbb{F}_p)$ and $\displaystyle |\text{GL}_2(\mathbb{F}_p)|=(p^2-1)(p^2-p)=p(p-1)^2(p+1)$. I forgot about the case when $\displaystyle p=3$, etc. Good catch.