Class equation of normal subgroups

Question

I am puzzled with a thought or a question regarding class equation and normality of subgroups.Consider the following situation.

Let $G$ be a finite group and $N\trianglelefteq G$.Let $G$ act on $N$ by conjugation which is allowed since $N$ is normal and let the representatives of the different equivalence classes be $n_1,n_2,...,n_k$ (excluding the equivalence class of identity).

Then, the equivalence class of $n_i$ is

$[n_i]=\{gn_ig^{-1} : g\in G\},i=1,2,..,k$

Then $N=\{e\}\displaystyle\cup_{i=1}^k[n_i]$

$\Rightarrow |N|=1+\sum_{i=1}^k |[n_i]| \quad (1)$

Now, let $N$ act on itself by conjugation.If $cl(n_i)=\{nn_in^- : n\in N\}(i=1,2,3,..,k)$ be the equivalence of $n_i$ under this action.Then I can see that

$cl(n_i)\subseteq [n_i]$ but can it happen that $cl(n_i)\subsetneq [n_i]$ for some $i$(i.e a proper subset) or if the question is put in other way

Will the class equation of $N$ be the same as $(1)$?

This may be a silly question but I am trying to keep my understandings clear.

I have seen a question with the same title here(Class equation of normal subgroup) but I don't think that answers my question.

Any suggestions or advice?

Thanks for your time.

Answer 1

Let's see an example, We'll take $$G=S_3=\{e,(12),(13),(23),(123),(132)\}$$ and

$$N=A_3=\{e,(123),(132)\}$$

We look at the action of $G$ on $N$ by conjugation and we get the equivalent classes

$$\{\{e\},\{(123),(132)\} \}$$

So here the class equation is :

$$3=|N|=1+|[(123)]|=1+2$$

We look at the action of $N$ on itself we get:

$$\{\{e\},\{(123)\},\{(132)\}\}$$

So here the class equation is:

$$3=|N|=1+cl([(123)])+cl(|[(132)])=1+1+1$$

So,the $n_i$ are the representatives of the equivalent classes for the action of conjugation of $G$ on $N$.

But the there could be more equivalence classes when you take the action of N on itself.

so in the example here, we have that $[(123)]=\{(123),(132)\}$ but $cl((123))=\{(123)\}$

So when you write down the class equation for the action of $N$ on itself, you cannot just take the $cl(n_i)$, where $n_i$ are the class representative of the action of $G$ on $N$. Because you are missing more equivalence classes for in the action of $N$ on itself.