Unique Sylow p-Subgroups

**olgashukina** · December 3rd 2011, 12:28 PM

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.

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

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

**Drexel28** · December 3rd 2011, 02:53 PM

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.

**Deveno** · December 3rd 2011, 04:13 PM

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.

**Drexel28** · December 3rd 2011, 05:15 PM

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.

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