Determine the number of subgroups of $\Bbb Z_p \times\Bbb Z_p$, where $p$ is prime. There are some answers online and we got one in our lecture. Unfortunately I have spent several hours trying to make sense of it and getting nowhere. I think it is mainly due to the fact of me being very very poor at groups of the type integers modulo $n$.
I should note that I understand that the problem boils down to finding all the cyclic subgroups of $\Bbb Z_p \times \Bbb  Z_p$ of order $p$.
Help would be much appreciated, thank you!
 A: A subgroup of $\mathbb{Z}_{p}\times\mathbb{Z}_{p}$ must have order dividing $p^2$ by Lagrange's theorem. Since $p$ is prime, the possible orders of subgroups of $\mathbb{Z}_{p}\times\mathbb{Z}_{p}$ are $1,p,p^2$. For $1,p^2$ there are only two subgroups of $\mathbb{Z}_{p}\times\mathbb{Z}_{p}$ with that order, namely $\{(e,e)\}$ and $\mathbb{Z}_{p}\times\mathbb{Z}_{p}$ respectively, where $e\in\mathbb{Z}_{p}$ is the identity element.
So now suppose that $A\leq \mathbb{Z}_{p}\times\mathbb{Z}_{p}$ with $|A| = p$. Then since $p$ is prime, $A$ must be cyclic, so there exists some element $(a,b)\in\mathbb{Z}_{p}\times\mathbb{Z}_{p}$ such that $A = \langle(a,b)\rangle $, so $(a,b)$ must have order $p$ in $\mathbb{Z}_{p}\times\mathbb{Z}_{p}$.
The converse is also true, i.e. if $(a,b)$ has order $p$ then the subgroup it generates has order $p$. So the set of subgroups of $\mathbb{Z}_{p}\times\mathbb{Z}_{p}$ of order $p$ is $\{\langle(a,b)\rangle\mid(a,b) $ has order $p$$\}$.
So now note that the elements of order $p$ of $\mathbb{Z}_{p}\times\mathbb{Z}_{p}$ are exactly the elements of the form $(a,b)$ where either $a\neq e$, or $b\neq e$ (or both $\neq e$). That is, they are exactly the elements of $\mathbb{Z}_{p}\times\mathbb{Z}_{p}\setminus \{(e,e)\}$, of which there are $p^2-1$. However, each element of order $p$ accounts for $p-2$ other elements of order $p$. You can partition the set of elements of order $p$ into equivalence classes under the equivalence relation $(a,b)\sim(c,d)$ iff there exists $t\in\mathbb{Z}$ such that $(a,b)^{t} = (c,d)$. Each equivalence class has $p-1$ elements (note the identity element of $\mathbb{Z}_{p}\times\mathbb{Z}_{p}$ is not in the set of elements of order $p$), and so the number of equivalence classes is $\frac{p^2-1}{p-1} = p+1$. So there are $p+1$ subgroups of order $p$.
Now adding to account for the subgroups $\{(e,e)\}$ and $\mathbb{Z}_{p}\times\mathbb{Z}_{p}$, we have that $\mathbb{Z}_{p}\times\mathbb{Z}_{p}$ has $p+3$ subgroups.
A: For Zp × Zp group order has p²
and  possible order of subgroup is 1,p,p²
and we get always one 1 oder subgroup and one p² order subgroup.
Now we have just find out no of subgroup of order p.
and no of subgroup of order p
= no of elements of order p / φ(p)
= (p²-1)/(p-1)
= p+1
:. total number of subgroup
= 1+1+p+1
=p+3
