# Generators of two groups with prime order $p$ already induce all the generators of the product group $G \times H$

Let $G = \langle g \rangle, H = \langle h \rangle$ be two cyclic groups (with $g \in G, h \in H$), both of them of order $p \in \mathbb{N}$, where $p$ is a prime number.

I now want to show that

$$\langle(g, h^0)\rangle, \langle (g, h) \rangle, \langle (g, h^2) \rangle, ..., \langle (g, h^{p-1})\rangle, \langle (g^0, h) \rangle$$

are already all of the subgroups of order $p$ of the group $G \times H$.

Now I must admit that I couldn't really get started so far. I can see that each of the subgroups of $G \times H$ listed above are indeed of order $p$. (Because $ord(g) = ord(h) = p$, and $ord(h^k), ord(g^k) ≤ p$ for $k \in \mathbb{N}$, and each of the generators has either $g^1$ or $h^1$ in one of it's components.)

Because $p$ is prime, we have that $g^k$ is a generator of $G$ aswell for all $k \in \mathbb{N}$, unless $g^k = e_G$, I believe. (Because for a generator $g$ of $G$ with order $n$, we have that $g^i$ is a generator iff $i$ and $n$ are co-prime and $g^i ≠ e_G$, I think?)

I don't really know how to continue from there in order to show that the list above already features all subgroups of $G \times H$ of order $p$, i.e. that there are no other ones than the ones given in the list above.

• An element of $G\times H$ is of the form $g^mh^n$ where both $m$ and $n$ can take values from $0$ to $(p-1)$. Unless $m=n=0$, all such elements are of order $p$. Excluding the case $m=n=0$ which corresponds to identity, all elements of $G\times H$ generate cyclic subgroups of order $p$. Consider the group $<gh^r>$. Since every non-identity element of $<gh^r>$ is a generator, the group $<gh^r>$ is same as the groups $<g^2h^{2r}>$, $<g^3h^{3r}>$ ... $<g^{p-1}h^{r(p-1)}>$. – vnd May 5 '16 at 13:37

$G\times H$ has $p^2$ elements. One of them is the identity, all the others have order $p$. An element of order $p$ is contained in a unique subgroup of order $p$, and each such subgroup contains $p-1$ elements of order $p$. It follows that there are $\frac{p^2-1}{p-1}=p+1$ subgroups of order $p$. As you have exhibited $p+1$ of them already, it suffices to check that they are distinct.
This is easy. For example, consider which elements of the form $(g,h^a)$ they contain. The first $p$ of them contain a unique such, with the value of $a$ always being different, whereas the last one doesn't even contain such an element.