Exercise 3, Section 1.5 of Hungerford’s Abstract Algebra

Question

Definition: Let $(G,\circ)$ be a group. A congruence relation on $G$ is an equivalence relation $\equiv$ on the elements of $G$ satisfying if $g_1\equiv g_2$ and $h_1\equiv h_2$, then $g_1\circ h_1\equiv g_2\circ h_2$ for all $g_1,g_2,h_1,h_2\in G$.

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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 $G$.

My attempt: $(\Rightarrow)$ Suppose $N$ is normal in $G$. Let $a\equiv_r b$ and $c\equiv_r d$. Then $ab^{-1}\in N$ and $cd^{-1}\in N$. We need to show $ac\equiv_r bd$, i.e. $ac(bd)^{-1}=acd^{-1}b^{-1}\in N$. Since $N$ is normal in $G$, we have $N=pNp^{-1}$, $\forall p\in G$. Take $p=a^{-1}$. So $N=a^{-1}Na$. Since $cd^{-1}\in N$, we have $\exists n\in N$ such that $cd^{-1}=a^{-1}na$. Multiply by $a$ on left side, $acd^{-1}=aa^{-1}na=na$. Multiply by $b^{-1}$ on right side, $acd^{-1}b^{-1}=nab^{-1}$. Since $ab^{-1}\in N$, we have $nab^{-1}\in N$. So $acd^{-1}b^{-1}\in N$ and $ac\equiv_r bd$. Thus $\equiv_r$ is congruence relation on $G$.

$(\Leftarrow)$ Suppose $\equiv_r$ is congruence relation on $G$. We show $aNa^{-1}\subseteq N$, $\forall a\in G$. Let $n\in N$. It’s easy to check, $a\equiv_r a$ and $n\equiv_r e$. Since $\equiv_r$ is congruence relation, we have $an\equiv_r a$. That is $ana^{-1}\in N$. Thus $aNa^{-1}\subseteq N$. Hence $N\lhd G$. Is my proof correct?

I’m uncertain about my use of definition of normal subgroup in $(\Rightarrow)$. Since there are several equivalent definition of normal subgroup, I assume there is more than one way to solve this problem.

Answer 1

Your argument is sufficient and fine. I would suggest the following computation $$ac(bd)^{-1}=acd^{-1}b^{-1}=a(cd^{-1})a^{-1}\cdot (ab^{-1})$$ which is an element in $N$ by the same reasoning as yours.

Answer 2

I don't see why you have doubts here; although, I guess, if you're new to this, it can be daunting: there's a lot of symbols to keep track of.

The proof seems fine to me.

As to the particular aspect of the proof you are concerned with: the definition of "normal" you use is, of course, perfectly legitimate; why wouldn't it be?