A normal subgroup is the union of conjugacy classes.

This is Exercise 2.6.5 of F. M. Goodman's "Algebra: Abstract and Concrete". I want to check my proof.

Exercise 2.6.5: Show that a subgroup (of a group) is normal if and only if it is the union of conjugacy classes.

My Attempt:

Let $N$ be a subgroup. Then

\begin{align} N\text{ is normal }&\Leftrightarrow \forall g\in G, N=gNg^{-1} \\ &\Leftrightarrow \forall n\in N \forall g\in G\exists m_{n, g}\in N, n=gm_{n, g}g^{-1} \tag{1}\\ &\Leftrightarrow N=\bigcup_{n\in N}\underbrace{\bigcup_{g\in G}\left\{gm_{n, g}g^{-1}\right\}}_{\text{conjugacy class of }n}\tag{2} \\ &\Leftrightarrow N=\bigcup_{n\in N}[n], \end{align} where $[n]$ is the conjugacy class of $n$.$\square$

Is this proof valid?

Thoughts:

I hope to make $(1)$ to $(2)$ (and back) more explicit.

• It would be a better proof if it consisted of English sentences rather than just a wall of symbols. – Henning Makholm Mar 7 '17 at 11:20
• @HenningMakholm True. – Shaun Mar 7 '17 at 11:21
• @HenningMakholm I wanted to use "if and only if" statements all the way. Presenting it like this seemed most efficient. – Shaun Mar 7 '17 at 11:22

If $N$ is a normal subgroup of group $G$ and $n\in N$ then $gng^{-1}\in N$ for every $g\in G$ or equivalently $[n]\subseteq N$ where $[n]:=\{gng^{-1}\mid g\in G\}$ is the conjugacy class of $n$.

This tells us that: $$N=\bigcup_{n\in N}[n]$$ If conversely $N$ is a subgroup of group $G$ that satisfies $N=\bigcup_{n\in N}[n]$ then it is immediate that $gng^{-1}\in N$ for every $n\in N$ and $g\in G$, so the conclusion that $N$ is a normal subgroup is justified.

• in your definition of normal subgroup you write $gng^{-1}$ gor every g$\in$N .i think it is not 'N' but 'G'.please check it out... – Cloud JR Sep 22 '18 at 7:46
• Elegant proof +1 – Cloud JR Sep 22 '18 at 7:48
• @CloudJR Thank you. You are right. I repaired now. – drhab Sep 22 '18 at 11:02