6
$\begingroup$

Question:

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.

Thanks

$\endgroup$
  • 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 $H$. $\endgroup$ – Tobias Kildetoft Oct 10 '13 at 7:38
2
$\begingroup$

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 conatained 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.

$\endgroup$
0
$\begingroup$

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 $G$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.