I want to prove that the group $(\mathbb{Z}/p^2\mathbb{Z})^\times\ltimes \mathbb{Z}/p\mathbb{Z}$ has a unique subgroup of index $p^2$.
I know that $(\mathbb{Z}/p^2\mathbb{Z})^\times$ is cyclic of order $p(p-1)$, hence by the Chinese remainder theorem isomorphic to $\mathbb{Z}/p\mathbb{Z}\times \mathbb{Z}/(p-1)\mathbb{Z}$, but this doesn't seem to get me much further.
My teacher sees this as a triviality, but I don't seem to see it.. Could someone provide any help?
The order of your group is $p^2(p-1)$, so every subgroup of index $p^2$ has order $p-1$. For $p=2$ the claim is clearly true as any group has a unique subgroup of order $1$. So suppose $p>2$.
Let $a\in(\Bbb{Z}/p^2\Bbb{Z})^{\times}$ and $b\in\Bbb{Z}/p\Bbb{Z}$. Computing powers of $(a,b)$ shows that $$(a,b)^2=(a^2,b(a+1)),\qquad (a,b)^3=(a^3,b(a^2+a+1)),$$ and in general, if $a\neq1$, for any positive integer $k$ we have $$(a,b)^k=\Big(a^k,b\sum_{i=0}^{k-1}a^k\Big)=\Big(a^k,b\tfrac{a^k-1}{a-1}\Big).\tag{1}$$ From this it easily follows that the order of $(a,b)$ equals the order of $a$.
Now pick some $a\in(\Bbb{Z}/p^2\Bbb{Z})^{\times}$ of order $p-1$ and pick any two distinct elements $b_1,b_2\in\Bbb{Z}/p\Bbb{Z}$. Then the expression in $(1)$ makes it easy to verify that for all $k$ we have $$(a,b_1)\neq(a,b_2)^k,$$ so $(a,b_1)$ and $(a,b_2)$ generate two distinct subgroups of order $p-1$.