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 :)

  • 6
    $\begingroup$ It would be a better proof if it consisted of English sentences rather than just a wall of symbols. $\endgroup$ Commented Mar 7, 2017 at 11:20
  • $\begingroup$ @HenningMakholm True. $\endgroup$
    – Shaun
    Commented 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
    Commented Mar 7, 2017 at 11:22
  • 1
    $\begingroup$ @Shaun sir ,very interesting question. $\endgroup$ Commented Jul 21, 2022 at 16:59
  • $\begingroup$ Thank you, @LostinSpace. $\endgroup$
    – Shaun
    Commented Jul 21, 2022 at 17:46

4 Answers 4


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.

  • 1
    $\begingroup$ 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
    Commented Sep 22, 2018 at 7:46
  • $\begingroup$ Elegant proof +1 $\endgroup$
    – Cloud JR K
    Commented Sep 22, 2018 at 7:48
  • 1
    $\begingroup$ @CloudJR Thank you. You are right. I repaired now. $\endgroup$
    – drhab
    Commented 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
    Commented Dec 20, 2019 at 10:00
  • $\begingroup$ I know this is an old post but the proof is not complete as stated. The fact that $gng^{-1}\in N$ for all $n\in N$ and $g\in G$ does not immediately imply that $gN=Ng$. It could be that $gNg^{-1}$ is a subset of $N$ for all $g\in G$, but not necessarily equal. This cannot happen, but that's only because conjugation by $g$ is a bijection on $G$ with inverse conjugation by $g^{-1}$. $\endgroup$ Commented May 20 at 21:43

A subset $Y$ of a set $X$ with a $G$ action is invariant if and only if it is a union of orbits. Here $X=G$, the action is $g \cdot h : = g h g^{-1}$. Btw, there is another action of $G\times G$ on $G$,

$$(g_1, g_2) \cdot g = g_1 g g_2^{-1}$$


The reverse direction is pretty straight-forward: If $N_0\triangleleft G$, then $x\in N_0$ if and only if $[x]\subset N_0$. Thus, $N_0$ must be a union of conjugacy classes.

Similarly so, the forward direction is also rather easy to see! Denote $N=\bigcup_{x\in H<G}[x]$ where $[x]=\{gxg^{-1}:g\in G\}$. We can quickly check the group axioms as follows:

  • Well-Defined Associative Binary Operator (inherited from group $G$): All we need to check in order to see that the group operator on $G$ induces an (associative) binary operator on $N$ is closure (i.e. $ax\in N$ for any $a,x\in N$). This is easy to check by fact that $N\supset[a][x]=\{g_aag_a^{-1}g_xxg_x^{-1}:g\in G\}\supset\{g(ax)g^{-1}:$$g\in G\}=[ax]$; hence, $[ax]\subset N$.

  • Identity: Since $H<G$, it follows immediately that $1\in N$ (since $H<N$).

  • Inverse: Using again the fact that $H<G$, it is immediate that $[x]\subset N$ implies $[x^{-1}]\subset N$ (if this isn't clear, just stare at the construction of $N$ for a while and it should be straight-forward).

  • Normal: Since $N$ is the union of conjugacy classes, we conclude normalcy. With this, we reach the desired conclusion.


The definition of a normal subgroup is that $gN=Ng$ for all $g\in G$, or equivalently $gNg^{-1}=N$ for all $g\in G$. Note the following pitfall: for a fixed $g\in G$, it is not true that $gng^{-1}\in N$ for all $n\in N$ implies that $gNg^{-1}=N$, it merely implies that $gNg^{-1}$ is a subset of $N$. For example, let $H=\mathbb Q$ under addition, let $G$ be the group obtained from $H$ by adding on top the automorphism $g:x\mapsto 2x$, and let $N=\mathbb{Z}$. Then $gNg^{-1}$ is a subset of $N$, but is not equal to it.

Suppose that $N$ is a union of conjugacy classes. This implies that, for all $g\in G$, $gNg^{-1}$ is a subset of $N$. Now we need that conjugation by $g$ is a bijection with inverse conjugation by $g^{-1}$. Conjugation by $g$ and $g^{-1}$ are bijections and the composition of the two of them, which is the identity, maps $N$ exactly onto $N$. Each of the two conjugations maps $N$ into $N$, and since the composition maps $N$ onto $N$, each conjugation maps $N$ onto $N$.

Another way to see this is to suppose that $x\in N\setminus gNg^{-1}$. There exists $y\in G$ such that $x=gyg^{-1}$, or equivalently $y=g^{-1}xg$. Since $x\in N$, $y\in N$, and so $x\in gNg^{-1}$, a contradiction.

But you do need this argument to see why $gNg^{-1}=N$, and not is just a subset of it.


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