Exercise 4, Section 1.5 of Hungerford’s Abstract Algebra

Question

Let $\backsim$ be an equivalence relation on a group $G$ and $N=\{a\in G\mid a\backsim e\}$. Then $\backsim$ is a congruence relation on $G$ if and only if $N$ is a normal subgroup of $G$ and $\backsim$ is congruence modulo $N$.

I have written a solution below. Exercise 4 section 5 is corollary of exercise 3 section 5. We only need to prove $\backsim =\equiv_r \,\operatorname{mod} N$, rest follows from exercise 3 section 5. Which states “Let $N$ be a subgroup of a group $G$. $N$ is normal in $G$ if and only if (right) congruence modulo $N$ is a congruence relation on $G$”.

Answer 1

Proof: $(\Rightarrow)$ Suppose $\backsim$ is a congruence relation on $G$. Clearly $N=\{a\in G\mid a\backsim e\}$ is equivalence class of $e$ and $N\leq G$. We show $\backsim =\equiv_r \,\operatorname{mod} N$. Let $(a,b)\in \backsim$. Then $a\backsim b$. Since $\backsim$ is equivalence relation, we have $a^{-1}\backsim a^{-1}$ and $b\backsim a$. Since $\backsim$ is a congruence relation, $a^{-1}b\backsim a^{-1}a=e$. That is $a^{-1}b\in N$. So $a\equiv_r b \,\operatorname{mod} N$. Thus $\backsim \subseteq \equiv_r \,\operatorname{mod} N$. Let $(a,b)\in \equiv_r \,\operatorname{mod} N$. Then $a^{-1}b\in N$ and $a^{-1}b\backsim e$. Since $\backsim$ is congruence relation, we have $a=ab^{-1}b\backsim eb=b$. So $a\backsim b$. Thus $\backsim \supseteq \equiv_r \,\operatorname{mod} N$. Hence $\backsim = \equiv_r \,\operatorname{mod} N$. By exercise 3 section 5, $N$ is normal in $G$.

$(\Leftarrow)$ Suppose $N\lhd G$ and $\backsim =\equiv_r \,\operatorname{mod} N$. By exercise 3 section 5, $\equiv_r \,\operatorname{mod} N$ is congruence relation on $G$.