Congruence between groups when a homomorphism is present

Question

Let $G$, $H$ be groups, and $\phi$ surjectiv homomorphism from $G$ to $H$. How can I show that if $\alpha$ is a congruence of $G$, then $$\phi(\alpha)=\{(\phi(a),\phi(b)) : (a,b) \in \alpha\} $$ is congruence of $H$?

Answer 1

By definition $\alpha$ is an equivalence relation compatible with the group operations (multiplication and inverse), and $\phi$ intertwines the group operations of $G$ and $H$. You can just put everything together and turn the handle to show that $\phi(\alpha)$ is a congruence on $\phi(G)$.

For instance, suppose $a_1,a_2,a_3,a_4 \in \phi(G)$ are such that $(a_1,a_2),(a_3,a_4) \in \phi(\alpha)$. Then $a_i = \phi(b_i)$ where $(b_1,b_2),(b_3,b_4) \in \alpha$. Since $\alpha$ is a congruence, $(b_1b_3,b_2b_4) \in \alpha$, so $(\phi(b_1b_3),\phi(b_2b_4)) \in \phi(\alpha)$. Since $\phi$ is a homomorphism, $\phi(b_1b_3) = a_1a_3$ and $\phi(b_2b_4) = a_2a_4$, so $(a_1a_3,a_2a_4) \in \phi(\alpha)$. A similar argument works for inverses.

I think in the group theory literature, it's not so common to talk about congruences, but rather normal subgroups (since a congruence on a group is just the equivalence relation formed by the cosets of a normal subgroup). In those terms, if $\alpha$ is the congruence associated to the normal subgroup $N$ of $G$, then $\phi(\alpha)$ is the congruence on $\phi(G)$ associated to the normal subgroup $\phi(N)$. Again, the proof doesn't require any imagination.