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 $p$.
Best Answer
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.