Interpreting the Problem Statement

[Bashyboy]

Homework Statement

Assume that $H$ is a normal subgroup of $G$, $\mathcal{K}$ is a conjugacy class of $G$ contained in $H$, and $x \in \mathcal{K}$. Prove that $\mathcal{K}$ is a union of $k$ conjugacy classes of equal size in $H$, where $k = |G : HC_G(x)|$

Homework Equations

The Attempt at a Solution

Okay. I need a little help interpreting this problem. Is the problem asking me to show that $\mathcal{K} = \bigcup_{i \in I } \mathcal{H}_i$ with $|I| = k$, where the $\mathcal{H}_i$ are the conjugacy classes formed by letting $H$ act on itself by conjugation, or are the $\mathcal{H}_i$ the conjugacy classes formed by letting $G$ act on itself by conjugation that are contained in $H$?

 

[fresh_42]

Can you write this in the original language, which ever it is? Is $\mathcal{K}$ the set of all $gHg^{-1}$ or what is meant here, i.e. conjugates of what? And what is a class, i.e. if it is obviously different from a subset, how can it be included in $H$? And last but not least, what is $x$? And which kind of grammar is prove that is? No subject anywhere near.

Do they mean $xHx^{-1} \subseteq H \,$?

 

[Bashyboy]

Can you write this in the original language, which ever it is? Is $\mathcal{K}$ the set of all $gHg^{-1}$ or what is meant here, i.e. conjugates of what? And what is a class, i.e. if it is obviously different from a subset, how can it be included in $H$? And last but not least, what is $x$? And which kind of grammar is prove that is? No subject anywhere near.

Do they mean $xHx^{-1} \subseteq H \,$?

Sorry. It should read "Prove that $\mathcal{K}$..." I just edited it. Besides that, I typed up the problem word for word.

 

[fresh_42]

This is strange, sorry.

$\mathcal{K}$ is a conjugacy class of $G$ ...

This lacks a specification. Conjugacy class of what? Of $\{e\}$, of $G$, or of $H$, or of something else. Conjugacy class of $G$ would be $gGg^{-1}=G$ which makes no sense.
The guess would be $H$, but this is not self-evident.

... contained in $H$, and $x \in \mathcal{K}$.

Then $x$ is a set of the form $g_xHg_x^{-1}$ and $x=g_xHg_x^{-1} \subseteq H$ which means $g_x \in N_G(H)$ the normalizer of $H$. Is there a reason not to say this right away? So the claim is $|N_G(H)| = |G : HC_G(g_x)|$, is that correct?

Edit: My bad, this doesn't make sense either as $N_G(H)=G$. So sorry, I obviously don't understand the question.