Finding the subgroups of cyclic groups Currently trying to work out all of the subgroups of $X \times Y$, $X = C_3 = \langle x \rangle$ and $Y = C_3 = \langle y \rangle$. 
I know that $X = \{1, x, x^2\}$ and $Y = \{1, y, y^2\}$. I also know that
$$X \times Y =  \{(1,1), (1,y), (1,y^2), (x,1), (x,y), (x,y^2), (x^2, 1), (x^2, y), (x^2, y^2)\}.$$
I came up with these subgroups:
\begin{align*}
H_1 &= \{(1,1), (1,y), (1,y^2)\} \\
H_2 &= \{(1,1), (x,1), (x^2,1)\} \\
H_3 &= \{(1,1), (x,y), (x^2,y^2)\} \\
H_4 &= \{(1,1), (x,y^2), (x^2,y)\} \\
H_5 &= \{(1,1), (1,y), (1,y^2), (x,1), (x,y), (x,y^2), (x^2, 1), (x^2, y), (x^2, y^2)\}
\end{align*}
I would just like to double check with someone that these are correct and that they're ALL of the possible subgroups. If not, I'd appreciate someone helping me out and pointing out/explaining my mistakes.

It's possible to be a bit systematic here, to make sure that we have them all.
$X\times Y$ has $9$ elements. That means any subgroup has either $1$, $3$ or $9$ elements. The orders of $1$ and $9$ are easily handled (as long as you remember them in the first place). That leaves $3$.
Any subgroup with $3$ elements will be cyclic. So just take each non-identity element in the group, and see which cyclic group it generates, and you're done.
There are $8$ non-identity elements, and they all have order $3$, which should make it $8$ order $3$ subgroups... Except not really. Since each order $3$ subgroup has $2$ potential generators, there is some overlap: $\langle(x, y)\rangle = \langle (x^2, y^2)\rangle$, for instance. This means that the $8$ different cyclic-subgroup-generators make a total of $4$ subgroups. And you have found $4$, so you have them all. Good job!
