Let $G$ be a group and $H$ a subgroup of $G$. A conjugacy class of an element $α∈G$ is the set $\def\CC{\mathop{\rm CC}}\CC(a)=\{ g^{-1} ag \mid g∈G\}$. Prove that $H$ is the union of conjugacy classes if and only if $H$ is normal in $G$.

My Answer:

Prove that if $H$ is normal in $G$ then $H$ is the union of conjugacy classes. Assume that $H$ is not the union of the conjugacy classes. This means that $H$ does not have an element that is of the form $g^{-1} ag$ where $a∈G$. Now, $H$ is normal in $G$, then $gH=Hg$ for all $g∈G$. Hence, for some $g∈G$ and $h∈H$ there exist $h'∈H$ such that $gh=h' g$. This shows that $h=g^{-1} h' g∈H$ which contradicts that $H$ does not have any element of the form $g^{-1} ag$, $a∈G$. Thus, it is proven that if $H$ is normal in $G$, then $H$ is the union of conjugacy classes.

Conversely, prove that if $H$ is the union of conjugacy classes then $H$ is normal in $G$. A property of subgroup $N$ being normal to group $M$ is that for all $m∈M$, $mNm^{-1}⊆N$. Must then show that for all $g∈G$, $gHg^{-1}⊆H$. Let $a∈ gHg^{-1}$, then $a=ghg^{-1}$ for some $h∈H$. Now, the element $$a=ghg^{-1}=(g^{-1} )^{-1} hg^{-1}∈H.$$ Thus, by the stated property, $H$ is a normal subgroup in $G$.

Therefore it is proven that $H$ is the union of conjugacy classes if and only if $H$ is normal in $G$.

Is my answer correct or do i need to modify it? Kindly state if needed.


  • 2
    $\begingroup$ $H$ not being the union of conjugacy classes does not mean that $H$ contains no element of the form you claim. All elements in $G$ have that form, hence so do all elements of $H$. $\endgroup$ Oct 10, 2013 at 7:38

2 Answers 2


You are on the right track. Try to be a bit more straightforward in your answer. For instance, assuming that $H$ is normal in $G$ and then prove directly that $H$ is a union of conjugacy classes. To achieve that, you need to find the conjugacy classes that make up $H$. Consider, just one element $h\in H$. Since $H$ is normal, you know that $g^{-1}hg\in H$ for all $g\in G$. Doesn't that mean the $CC(h)$ is contained in $H$? So, you found a conjugacy class that is fully contains in $H$ and it contains the element $h$. But $h$ was arbitrary, so it is true in general that if $h\in H$ then its conjugacy class $CC(h)$ satisfies $h\in CC(h)\subseteq H$. Now, what does that tell you about $$\bigcup _{h\in H}CC(h)=?$$

Formalize carefully the arguments above and you'll have a proof of the first direction. You can then finalize the proof in the other direction.


think of $G$ acting on itself (as a set) via its inner automorphisms. the conjugacy classes are the minimal (non-empty) stable sets under the operation of the whole of $G$, and are disjoint.

let $C$ be any conjugacy class and $N$ a normal subgroup. then

$$ (C \cap N)^G = C^G \cap N^G = C \cap N $$ this set is thus stable under $G$ and must therefore (by minimality of $C$) either be empty or be a full conjugacy class, i.e. $C \subset N$. any conjugacy class either belongs to $N$ or is disjoint from it.

contrariwise, if $M$ is a subgroup, then if it is made up of complete conjugacy classes it is stable under all inner automorphisms of $G$ i.e. it is normal in $G$.


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