Prove $H\circ N = H\cap N$. Where $H$ and $N$ are two subgroups of a group $G$. I was trying to prove another theorem where I thought the above result could be helpful and started trying to prove it.
I am not sure whether the above statement is true or not, but I am unable to prove it. I'll be very thankful if somebody can help me out.
[NOTE: $H\circ N = \{h\circ n|h\in H, n\in N\}$].

Thanks to Mindlack, the doubt is now clear, and it was very silly.
$H\circ N$ as defined above should contain all elements of $H\cup N$, and only in the case $H=N$, $H\circ N = H\cap N$ is true.
