begin{proof}
Let $g_1, g_2 in G$. Then $theta(g_1) = Ng_1$ and $theta(g_2) = Ng_2$. \
begin{equation*}
begin{split}
text{ Now, } theta(g_1g_2) &= Ng_1g_2 \
& = Ng_1 Ng_2 \
& = theta(g_1) theta(g_2)\
end{split}
end{equation*}
$therefore theta$ is a group homomorphism. $theta$ is clearly onto.
end{proof}