Groups of prime squared order 
I have done part $a$ and $b$, stuck on $c$. I don't know what to do here. I know that since $G$ has order $p^2$, any  element must have order $1, p$ or $p^2$ so any generator must have one of those orders, but I don't know how to reason any further..
 A: So if you have no elements of order $p^2$, then all your elements have order $p$. Hence every group you have is cyclic of order $p$. Since they meet at the identity, there is a total of $x(p-1)+1=p^2$ elements. $x$ is what?
A: The following is, imo, a rather nice though perhaps not so used way to answer (c): first, if $\;G\;$ is cyclic there's nothing to prove as a finite cyclic group has one unique subgroup of each and every divisor of its order, so let us assume $\;G\;$ is not cyclic.
(1) Prove $\;G\cong C_p\times C_p\;$ , with $\;C_p=$ the cyclic group of order $\;p\;$ .
(2) Under the "scalar product" $\;k\cdot g:=g^k\;,\;\;g\in G\;,\;\;k\in \Bbb F_p:=\Bbb Z/p\Bbb Z\;$ , prove that $\;G\;$ is a vector space over the field $\;\Bbb F_p\;$ (note that as groups, $\Bbb F_p\cong C_p\;$ , of course)
(3) Now show that "vector subspace" and "subgroup" are exactly the same in this case.
(4) Finally, count how many $\;1-$dimensional subspaces $\;G\;$ has...
A: Here we know the group itself has order $p^{2}$, so if it is cyclic then you have a unique order $p$ subgroup. Otherwise pick up another element $b$ not in the $p$-group $A$ we already chosen, and we may argue that $|b|=p$ . Since you already proved that the intersection of $\langle b \rangle\cap A=e$, you can show $G=A\times \langle b\rangle$. So the group must be isomorphic to $\mathbb{Z}/p\mathbb{Z}\times \mathbb{Z}/p\mathbb{Z}$. Counting multiplicity of subgroups generated by pair $(x,y)$ and you find it is $p+1$. 
