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 of a group $G$. 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.
Please help 🙂
Best Answer
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.