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I am trying to solve the following:

Let $G$ a group and $N$ a normal subgroup of $G$ with index a prime $p$. Consider $C$ a conjugacy class of $G$, which is contained in $N$. Prove that either $C$ is a conjugacy class of $N$ or it is a union of $p$ distinct conjugacy classes of $N$.

[EDIT] Following the suggestion of @stewbasic I have done the following:

After denoting by $X^g=\{g^{-1}xg, x\in X\}$ and by $B$ the set of conjugacy classes of $N$ we can define the action $\cdot:G \times B \to B$ by $g\cdot A= A^g$. It is not that diffcult, but also not completely trivial, to verify that this is in fact an action and that this is trivial over $N$, so we can actually "define" the same action now of $G/N$ on $B$. With this new action I know now that for each $A \in B$ the $|Orb_{G/N}(A)|$ will be either $1$ or $p$ by the Orbit Stabilizer Theorem. If this is $1$ then all other orbits will have the same size and if that is $p$ we will have only one orbit. I think the problem would be done if $C$ given in the statement was a conjugacy class of $N$ but it is a conjugacy class of $G$. Am I missing something?

The thing is: I am studying for my PhD qualyfing exam and I have a lot of difficulties with algebra since I don't like it much. So, I wouldn't like any complete answer because although it is probably a very easy problem, I wanna learn something by thinking of it. So, I would be very thankful if I could have some hints about how to approach this problem or at least some comments about whether my first two approaches could take me somewhere or not!

Thank you so much, friends!

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  • $\begingroup$ do you know that conjugating a normal subgroup N you only get N again? $\endgroup$ – janmarqz Jul 24 '17 at 23:30
  • $\begingroup$ I'm confused. If you conjugate a normal subgroup, what you get is the same group again. So you're saying $C$ is either $N$, or a union of a bunch of $N$'s, which is just $N$. Do you maybe mean cosets or something? $\endgroup$ – Alfred Yerger Jul 24 '17 at 23:31
  • $\begingroup$ @Alfred Yerger C is either a conjugacy class of N or a Union of $p$ conjugacy classes of N. $\endgroup$ – bttmbrcelo Jul 24 '17 at 23:33
  • $\begingroup$ @janmarqz I know :| $\endgroup$ – bttmbrcelo Jul 24 '17 at 23:37
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If $X\subseteq G$ and $g\in G$, let $X^g=\{g^{-1}xg\mid x\in X\}$. Let $B$ denote the set of conjugacy classes of $N$.

Hint: Show that $C\mapsto C^g$ defines an action of $G$ on $B$. Show that this action is trivial on $N$, so it induces an action of $G/N$. What are the possible sizes of orbits under this action?

Finally suppose $C$ is a conjugacy class of $G$ contained in $N$. Pick any $c\in C$ and let $C'\in B$ be the conjugacy class of $N$ containing $c$. Show that $C$ is exactly the union of the orbit of $C'$ under the above action.

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  • $\begingroup$ Thank you very much! I am gonna try it now. $\endgroup$ – bttmbrcelo Jul 24 '17 at 23:58
  • $\begingroup$ I am gonna write somethings here, can you check if this is actually what you were thinking of? $\endgroup$ – bttmbrcelo Jul 25 '17 at 19:08
  • $\begingroup$ I started defining the action $\cdot: G \times B \to B$ by $g\cdot C = C^g$ here $B$ is the set of conjugacy classes of $N$. I checked that this is in fact an action and that this is trivial over $N$ and then I could define the "same" action but now of $G/N$ on $B$. In that case, by the Orbit-Stabilizer theorem we know that the size of the orbits can be either $p$ or $1$. I am with a problem to conclude since I dont know how to bring the initial conjugacy class, which was of $G$, to the game. Can you help-me? $\endgroup$ – bttmbrcelo Jul 25 '17 at 19:13
  • $\begingroup$ @bttmbrcelo I added another hint. $\endgroup$ – stewbasic Jul 25 '17 at 21:33
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You have to prove that number of conjugacy classes in N equals to
$|G : NS(x)|$, where $S(x)$ is the centralizer of element $x$ of $C$ in $G$. To do this show that number of conjugacy classes in $N$ equal to:

$|G : S(x)|/|N:N \cap S(x)| = |G:NS(x)|\times|NS(x):S(x)|/|N:N \cap S(x)| = |G:NS(x)|$.

(The proof of $|NS(x):S(x)| = |N:N \cap S(x)|$ is very similar to the proof of Second Isomorphism Theorem).

Considering that $p$ is prime, $NS(x)$ is either equal $N$ or $G$.

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