# Exercise 3, Section 1.5 of Hungerford’s Abstract Algebra

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 $$g_1,g_2,h_1,h_2\in G$$.

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.

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.

• Thank you so much for the answer. Your proof is concise. Jun 8 at 13:04

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?

• Thank you so much for the answer. Yaa I’m new to normal subgroup topic, so I’m feeling bit uncomfortable with my proofs. Jun 8 at 13:10
• You're welcome. If you think my answer is helpful, @user264745, then please don't forget to upvote it! Jun 8 at 13:11
• Yup upvoted. By the way I solved next exercise which is similar to this problem. I’m pretty confident solution of that problem is correct because solution of this problem is correct. I want to post that solution as answer. I remember you told me to add why given problem is relevant to the community. Next problem is “$\backsim$ is a congruence relation on $G$ if and only if $N$ is a normal subgroup of $G$ and $\backsim$ is congruence modulo $N$“. IMO next problem is corollary of exercise 3 section 5. I don’t know how this problem is relevant to the community. Can you please help me with that? Jun 8 at 14:02
• It's not so much about the relevance to the community as it is about helping us help you, @user264745. If you explain why it is you think it's a corollary, then I think that's sufficient context for here for people to want to help you. Good luck! Jun 8 at 16:31