# 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 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.

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

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... Commented Sep 22, 2018 at 7:46
• Elegant proof +1 Commented Sep 22, 2018 at 7:48
• @CloudJR Thank you. You are right. I repaired now. Commented Sep 22, 2018 at 11:02
• How will apply this result, to determine a subgroup of given order is normal, if the class equation of the group is given Commented Dec 20, 2019 at 10:00
• 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 $g^{-1}$. 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 $$N=\bigcup_{x\in H where $$[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 $$H, it follows immediately that $$1\in N$$ (since $$H).

• Inverse: Using again the fact that $$H, it is immediate that $$[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.