Number of conjugacy classes of elements of order $7$ in a group of order 168

abstract-algebrafinite-groupsgroup-theorysimple-groupssylow-theory

Let $G$ be a simple group of order $168$ (Here, we don't assume we know there is a unique such group). Compute the number of conjugacy classes of elements of order $7$ in $G$ [Hint: Consider Sylow $3$-subgroup].

I already proved that $n_7 =8$ so there are total $48$ many order $7$ elements in $G$. As hint suggested, I considered Sylow $3$-subgroup. First, I think I can't determine $n_3$ nor $n_2$. Any idea for this problem?

Best Answer

Hints.

  1. Since $n_7=8$ we have $|G:N_G(P_7)|=8$ and $|N_G(P_7)|=21$.

  2. There are exactly two conjugacy classes of elements of order $7$ in group $H=N_G(P_7)$.

  3. Elements of order $7$ from different conjugacy classes of group $H$ are not conjugate in group $G$.

Related Question