Using class equation to show that intersection is nontrivial

[Mr Davis 97]

Homework Statement

Let $G$ be a finite p-group and $Z(G)$ its center. If $N \not = \{e\}$ is a normal subgroup of $G$, prove that $N\cap Z(G) \not = \{e\}$.

Homework Equations

The Attempt at a Solution

Since $N$ is a normal subgroup we can let $G$ act on $N$ by conjugation. In a manner similar to the case when $G$ acts on itself, we can construct the following class equation. Let $n_1,\dots,n_r$ be representatives of the orbits of this action not contained in $N\cap Z(G)$. Then $$|G| = |N\cap Z(G)| + \sum_{i=1}^{r}[G : \operatorname{Stab}_G(n_i)].$$ Since some prime $p$ divides $|G|$ and $[G : \operatorname{Stab}_G(n_i)]$ for all $i\in [1,r]$, it follows that $p$ divides $|N\cap Z(G)|$. Hence $N \cap Z(G) \not = \{e\}$. QED

My main question is have I explained in enough depth how I obtain the class equation that I got? Do I need to show in a rigorous way that $$|G| = |N\cap Z(G)| + \sum_{i=1}^{r}[G : \operatorname{Stab}_G(n_i)],$$ or is what I have written sufficient?

 

[fresh_42]

I think it is sufficient.

The stabilizer formula is usually formulated for a single element. So if you like, you can show how the single element version $|G|=|G.x|\cdot |C_G(x)|=|G:C_G(x)|\cdot |C_G(x)|$ transforms into a version for entire $N$. You used that different orbits are either disjoint or equal to obtain the sum, and that all fix points of $N$ are central. But your version is o.k.