subgroup of group of order $p^2$ Let $p\ge5 $ be a prime. Then

$\mathbb{F_p\times\mathbb{F_p}}$ has atleast five subgroups of order p.

Every subgroup of $\mathbb{F_p\times\mathbb{F_p}}$ is of the form ${H_1\times{H_2}}$ where ${H_1 , {H_2}}$ are subgroups of $\mathbb{F_p}$.

Every subgroup of $\mathbb{F_p\times\mathbb{F_p}}$ is an ideal of the ring $\mathbb{F_p\times\mathbb{F_p}}$

The ring $\mathbb{F_p\times\mathbb{F_p}}$  is a field.


1...true .Let p=5 then subgroup generated by (1,0) , (2,0) , (3,0) , (4,0) , (0,1),....are all subgroups of order 5.
2...not true consider subgroup generated by (x,x) which cannot be written as ${H_1\times{H_2}}$.
3...I am not able to conclude for this option
I know that ideals of ring corresponds to  normal subgroups of corresponding group . $\mathbb{F_p\times\mathbb{F_p}}$ is abelian group so all subgroups are normal so can i say that all subgroups are normal subgroups and hence ideal ?
4...not true because in Fp×Fp  (1,0)(0.1)=(0,0) hence  it is not integral domain   hence it is not field.
(1) is true.  Consider any element $x(k,n) = (nk\pmod p) \times k$ with $0\leq n&lt;p$ and $k\neq 0\pmod p) $. It is easy to see that $x(k,n)$ gnerates a cyclic subgroup of $\Bbb{Z}_p \times \Bbb{Z}_p$.  It is also easy to show that the group generated by  $x(k,m)$ does not contain $x(k,n)$. Thus there are $p$ distinct subgroups of this form. 

(2) is false. Consider the subgroup $\{ k \times k\}$ with $0\leq k&lt;p$. This is not of the form $H_1 \times H_2$
(1) is true.  Consider any element $x(k,n) = (nk\pmod p) \times k$ with $0\leq n&lt;p$ and $k\neq 0\pmod p) $. It is easy to see that $x(k,n)$ gnerates a cyclic subgroup of $\Bbb{Z}_p \times \Bbb{Z}_p$.  It is also easy to show that the group generated by  $x(k,m)$ does not contain $x(k,n)$. Thus there are $p$ distinct subgroups of this form. 

(2) is false. Consider the subgroup $\{ k \times k\}$ with $0\leq k&lt;p$. This is not of the form $H_1 \times H_2$