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?


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

Please help :)

  • 4
    $\begingroup$ It would be a better proof if it consisted of English sentences rather than just a wall of symbols. $\endgroup$ Mar 7, 2017 at 11:20
  • $\begingroup$ @HenningMakholm True. $\endgroup$
    – Shaun
    Mar 7, 2017 at 11:21
  • $\begingroup$ @HenningMakholm I wanted to use "if and only if" statements all the way. Presenting it like this seemed most efficient. $\endgroup$
    – Shaun
    Mar 7, 2017 at 11:22
  • 1
    $\begingroup$ @Shaun sir ,very interesting question. $\endgroup$ Jul 21, 2022 at 16:59
  • $\begingroup$ Thank you, @LostinSpace. $\endgroup$
    – Shaun
    Jul 21, 2022 at 17:46

1 Answer 1


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.

  • $\begingroup$ 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... $\endgroup$
    – Cloud JR K
    Sep 22, 2018 at 7:46
  • $\begingroup$ Elegant proof +1 $\endgroup$
    – Cloud JR K
    Sep 22, 2018 at 7:48
  • 1
    $\begingroup$ @CloudJR Thank you. You are right. I repaired now. $\endgroup$
    – drhab
    Sep 22, 2018 at 11:02
  • $\begingroup$ How will apply this result, to determine a subgroup of given order is normal, if the class equation of the group is given $\endgroup$
    – sabeelmsk
    Dec 20, 2019 at 10:00

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