A Question Based on Properties of Direct Product of Fields I am unable to solve this particular problem in abstract algebra:

Let $p\geq$ 5 be a prime and $\mathbb{F}_p $ be field of p elements. Then is following statement true or false: $\mathbb{F}_p $ × $\mathbb{F}_p $ has at least five subgroups of order $p$.

I think it is false as I think subgroup of $\mathbb{F}_p$ ×$\mathbb{F}_p$  must be of form   $H_{1} $ × $H_{2}$ , where $H_{1} $, $H_{2} $ are subgroups of $\mathbb{F}_p $ and using Lagrange theorem I get only four subgroups of $\mathbb{F}_p$ ×$\mathbb{F}_p$.
But the answer is that it's true for which I have no clue.
 A: The geometric explanation of the other post is fine, but here's an easy method to find it in $2$-space, and indeed $n$-space.
There are $p^n$ points in an $\mathbb{F}_p$-vector space of dimension $n$. There are therefore $p^n-1$ non-zero points. There are $p-1$ non-zero points on a line, and no two lines share non-zero points (because a line is all scalar multiples of a point). Thus there are
$$ \frac{p^n-1}{p-1}$$
lines in $n$-dimensional space. If $n=2$, this yields $p+1$.
A: Here is an idea: draw a $5\times 5$ grid with $25$ dots representing $\Bbb{F}_5\times \Bbb{F}_5$. To find subgroups, you should notice that a subgroup of $\Bbb{F}_5\times \Bbb{F}_5$ of order $5$ is nothing more than a $1-$dimensional vector subspace of this $2-$dimensional $\Bbb{F}_5-$vector space. So, you just need to find $≥5$ lines through the origin. This is pretty easy to do: for instance $\ell_1=\langle(0,1)\rangle,\ell_2=\langle(1,1)\rangle,\ell_3=\langle(2,1)\rangle,\ell_4=\langle(3,1)\rangle,\ell_5=\langle(4,1)\rangle$. Actually, there is one more which is $\ell_6=\langle(1,0)\rangle$.
You can carry out this reasoning for all $p$ to demonstrate that there are $p+1$ lines through the origin of $\Bbb{F}_p\times \Bbb{F}_p$ and hence that there are $\ge 5$ subgroups of size $p$ as long as $p\ge 5$.
As a note: this demonstrates that the projective space $\Bbb{P}(\Bbb{F}_p^2)$ has cardinality $p+1$.
