Let $p$ be a prime. Find number of subgroups of order $p$ in $\mathbb{Z}_p^n = \mathbb{Z}_p\times\cdots\times\mathbb{Z}_p$, '$n$' times.
Consider any element $g\ne 1 \in \mathbb{Z}_p^n$, then order of $g$ is $p$. Also, order of $\mathbb{Z}_p^n$ is $p^n$, hence there are exactly $p^n-1$ elements of order $p$ in $\mathbb{Z}_p^n$. Now i want to parition these elements into groups of order $p-1$. How do I show from here that number of such subgroups will be $$\frac{p^n-1}{p-1}$$
Each element $g$ of order $p$ generates a subgroup $\{g^1,...,g^p\}$ of order $p$ containing exactly $p-1$ elements of order $p$. Here $g^p=e$, the identity elemnt of the group.
If $a,b$ has order $p$ and the two subgroups $\{a^1,....,a^p\},\{b^1,...,b^p\}$ contains some element $x$ other than identity of the group then, $a^m=x=b^n$ for some $m,n\in \{1,....,p-1\}$.
Suppose, $x^i=x^j$ for some $i,j\in \{1,....,p-1\}$. Then, $a^{mi}=a^{mj}\implies a^{m(i-j)}=e\implies p\big|(mi-mj)\implies i=j$ as $1\leq m\leq p-1$. So that, $x^1,...,x^{p-1}$ are all distinct elements of both the subgrous $\{a^1,....,a^p\},\{b^1,...,b^p\}$.
Hence, $\{a^1,....,a^p\}=\{x^1,...,x^p\}=\{b^1,...,b^p\}$. That's if two subgroup of order $p$ intersect non-trivially then they are actually equal.