Clarification on Proof that "for $p$ a prime, the elementary abelian group of order $p^2$ has exactly $p+1$ subgroups or order $p$" I'm reading in Dummit's Abstract Algebra the example (3) p. 156, where they prove the claim "for $p$ a prime, the elementary abelian group of order $p^2$, $E$ (i.e. $Z_{p} \times Z_{p}$, where $Z_{p}$ is the cyclic group of order $p$) has exactly $p+1$ subgroups or order $p$".
They start proving this by saying that, since obviously each nonidentity element of $E$ has order $p$ (since it happens for every element of $Z_{p}$), then they generate a cyclic subgroup of order p. Noting that each of this subgroups has exactly $p-1$ generators (since $p$ is prime) and that by Lagrange's Theorem distinct subgroups of order $p$ intersect trivially, they deduce there are $\frac{p^2-1}{p-1}=p+1$ subgroups of order $p$.
I agree with the fact we have found $p+1$ subgroups of order $p$, but I don't know why they suppose they are all the subgroups of order $p$.
 A: Each of the $(p+1)$ subgroups has order $p$, and they intersect trivially.  Thus we count $(p+1)(p-1)+1=p^2$ elements in the group.
A: There are $p^2-1$ elements of order $p$.
Each of the non-intersecting subgroups contains $p-1$ of these elements.
Therefore there are precisely $\frac{p^2-1}{p-1}=p+1$ such subgroups.
A: An alternative is to think of this in terms of linear algebra. The elementary abelian group of order $p^n$ is an $n$-dimensional vector space over $\mathbb{F}_p$; the subgroups of order $p$ are exactly the $1$-dimensional subspaces. Each one dimensional subspace contains $p$ points, and each nonzero point can function as a basis for the subspace. Moreover, the intersection of two distinct $1$-dimensional subspaces must be trivial (dimension $0$) by dimensional considerations.
Thus, each nonzero element generates a one-dimensional subspace, and every one-dimensional subspace is thus obtained,  and we are counting each of them $p-1$ times. Thus, the number of distinct $1$-dimensional subspaces of the elementary abelian group of order $p^n$ is $\frac{p^n-1}{p-1}$.
One can easily generalize this for $k$-dimensional subspaces, i.e. subgroups of order $p^k$: you need to select a linearly  independent subset with $k$ elements. The first one can be picked arbitrarily, $p^n-1$ choices; the next one must be outside the one-dimensional subspace that first generates, $p^n-p$ choices; the next $p^n-p^2$ choices, and so on until the $k$th one has $p^n-p^{k-1}$  choices.  But then we have to account for the overcount. How many distinct ordered bases does a $k$-dimensional space over $\mathbb{F}_p$ has? Well,  using the same counting technique we see that we can pick  an ordered basis in
$$(p^k-1)(p^k-p)\cdots(p^k-p^{k-1})\quad\text{ways.}$$
Thus, the number of $k$-dimensional subspaces is
$$\frac{(p^n-1)(p^n-p)\cdots(p^n-p^{k-1})}{(p^k-1)(p^k-p)\cdots(p^k-p^{k-1})}.$$
In particular, for $n=2$ and $k=1$ we get
$$\frac{p^2-1}{p-1} = p+1.$$
