Number of Subgroups of $C_p \times C_p$ and $C_{p^2}$ for $p$ prime As the title says, I am interested in finding all subgroups of $C_p \times C_p$ and $C_{p^2}$ for $p$ prime.
We did not cover the Sylow-theorem so far in the lecture.
What I noticed so far: 
As $C_p$ is of order $p$, the elements can only have order $p$ or $1$ due to Lagrange's theorem. There is only one element of order 1, namely the neutral element. Hence there are $p^2-1$ elements in $C_p \times C_p$ with order $p$. 
Because $p$ is prime, $C_p$ is cyclic and hence there exists an element, call it $a$ which is of order p and generates the whole group. All other powers of $a$ are generating $C_p$ as well, so $(a^i,1)$ for $i=1 \dots p-1$, generate one subgroup that is not trivial. 
Using the same argumentation for the second factor, I conclude that there are $2$ nontrivial subgroups and $2$ trivial ones. 
Consider the case $p=2$ now.
$<(a,a)>$ is one as well, hence $5$ subgroups. But I am stuck with what happens for $(a^i, a^j)$ for $i \neq j$ and $i, j >0$ for general $p$ prime. Could you post some hints, please? 
For $C_{p^2}$, we know that there is at least one element, call it $c$. This generates the whole group. According to Lagrange, there can only be elements that are either of order $p^2, p$ or $1$. All $p^2$ elements will generate the whole group, so they are quite uninteresting. 
For $p=2$ again, $<2>$ is another subgroup, nontrival of order $p$. For this case, there are in total $3$ subgroups ($2$ trivial ones and $<2>$). I cannot find any meaningful generalisation. Any help is greatly appreciated.
 A: I decided to work out a solution a bit more. There's still some small gaps for you to fill in; let me know if they pose any trouble.
Let's start with $C_p \times C_p$. The easiest subgroups are the cylic ones, i.e. the subgroups which are generated by one element. If this element is $(1, 1)$, then you have the trivial subgroup. If this element is $(a, b)$ where $a$ and $b$ are not both $1$, then it generates the subgroup $\lbrace (1, 1), (a, b), (a^2, b^2), \ldots, (a^{p-1}, b^{p-1}) \rbrace$. Contrary to what you said, there are in general more than two of these subgroups. Indeed, for fixed $a, b \in C_{p}$ with $b \neq 1$ and $a \neq 1$, the elements $(a, 1), (a, b), (a, b^2), \ldots, (a, b^{p-1})$ and $(1, b)$ all generate different subgroups of $C_p \times C_p$ (check this!), so there are at least $p+1$ non-trivial subgroups of $C_p \times C_p$. Check for yourself that every cyclic subgroup of $C_p \times C_p$ is in fact generated by one of the mentioned $p$ elements. It then follows easily that these $p+1$ subgroups are exactly the non-trivial subgroups of $C_p \times C_p$: if you have another non-trivial subgroup, then it must strictly contain a cyclic subgroup, such that it is not proper by Lagrange's theorem (for you to check: why?).
I mostly covered the case of $C_{p^2}$ in the comments. Fix a generator $x$ of $C_{p^2}$, then $C_{p^2} = \lbrace 1, x, x^2, \ldots, x^{p^2 - 1} \rbrace$. An element $c^k$ with $k \in \lbrace 1, \ldots, p^2 -1 \rbrace$ has order $p$ or $p^2$. If it has order $p$, then $c^{kp} = 1$, such that $kp$ is a multiple of $p^2$ and thus $k$ is divisible by $p$. The only elements of order $p$ are the $p - 1$ elements $c^{pm}$ where $m \in \lbrace 1, \ldots, p - 1 \rbrace$ (why do they have order $p$?). Any proper subgroup of $C_p \times C_p$ can not contain any element of order $p^2$ (why not?) and any non-trivial subgroup must contain an element of order $p$, but then it must contain all elements of order $p$. Hence, there is only one proper, non-trivial subgroup of $C_{p^2}$ (namely the set of all elements of order $p$ plus $(1, 1)$) and it is isomorphic to $C_p$.
A: Every element of order $p$ in $G=C_p\times C_p$, and there are $p^2-1$ of them, generates a cyclic subgroup of order $p$, and every such subgroup has $p-1$ generators. This implies that there are $\frac{p^2-1}{p-1}$ subgroups of order $p$, that is, $p+1$. As there are also the trivial group, and the whole group (and no others, in view of Lagrage's theorem), the number of subgrups of $G$ is $p+3$.
On the other hand, a cyclic group $C_n$ has exactly one subgroup for each divisor of $n$, so $C_{p^2}$ has three subgroups, corresponding to the divisors $1$, $p$ and $p^2$ of $p^2$.
A: the error is human; the author has not written p ^ 2-2 nowhere in the text, thank you.
For sub-Gropes not trivials of C_pxC_p there are exactly the number of element of order p divided by p-1 therefore p ^ 2-1 / p-1 = p + 1. therefore the nompbre of subgroups  (included trivial groups) is p + 2. Result depends only on the fact that p is a prime number, it is not linked to the particularity of the number p=2.
For subgroup generated by the elements indicated in the text just go to their determination and identify those that coincide or seek those who remain.
