# Math Help - Unique Sylow p-Subgroups

1. ## Unique Sylow p-Subgroups

If p is an odd prime, prove the following.
a) If G is a group of order $(p-1)p^2$, then G has a unique Sylow p-subgroup.
b) There are at least four groups of order $(p-1)p^2$which are pairwise nonisomorphic.

I know little about Sylow's subgroups.

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

$n_i\equiv 1 \ (\text{mod} \ i)$

2. ## Re: Unique Sylow p-Subgroups

Originally Posted by olgashukina
If p is an odd prime, prove the following.
a) If G is a group of order $(p-1)p^2$, then G has a unique Sylow p-subgroup.
You know that the number of Sylow $p$-subgroups of $G$ divides $p-1$ and is equivalent to $1$ modulo $p$. Now, tell me, how many numbers strictly less than any given $n$ are equivalent to $n$ modulo $1$?

b) There are at least four groups of order $(p-1)p^2$which are pairwise nonisomorphic.
Ok, once you take care of the obvious abelian ones, the important observation is that $p-1\mid |\text{Aut}(\mathbb{Z}_{p^2})|=\varphi(p^2)=p(p-1)$ and so there exists non-trivial ____ products.

3. ## Re: Unique Sylow p-Subgroups

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

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

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

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

4. ## Re: Unique Sylow p-Subgroups

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

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

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

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