Question

Prove Theorem 9.25.

Short answer

The theorem 9.25 is proved.

Step-by-step solution

Statement

Let H and A be subgroups of a finite group G. The number of distinct H-conjugates of A (that is, the number of elements in the equivalence class of A under H-conjugacy) is $\left[\mathrm{H}:\mathrm{H}\cap \mathrm{N}\left(\mathrm{A}\right)\right]$ and therefore, divides $\left|\mathrm{H}\right|$ .

Define a map

Let S be the set of right cosets of $\mathrm{H}\cap \mathrm{N}\left(\mathrm{A}\right)$ in H and T let be the set of H- conjugacy classes of A in G.

Define a map $\mathrm{f}:\mathrm{S}\to \mathrm{T}$such that $\mathrm{f}\left(\left(\mathrm{H}\cap \mathrm{N}\left(\mathrm{A}\right)\right)\mathrm{x}\right)={\mathrm{x}}^{-1}\mathrm{Ax}$.

Here, $\left|\mathrm{S}\right|=\left[\mathrm{H}:\mathrm{H}\cap \mathrm{N}\left(\mathrm{A}\right)\right]$and $\left|\mathrm{T}\right|=$Number of H- conjugacy classes of A in G.

Prove that the map is bijective

From equivalences, it is clear that f is well defined and injective.

Find surjectivity as:

$$\begin{gathered} \left(\mathrm{H}\cap \mathrm{N}\left(\mathrm{A}\right)\right)\mathrm{x}=\left(\mathrm{H}\cap \mathrm{N}\left(\mathrm{A}\right)\right)\mathrm{y} \\ \iff {\mathrm{xy}}^{-1}\in \mathrm{H}\cap \mathrm{N}\left(\mathrm{A}\right) \\ \iff \left({\mathrm{xy}}^{-1}\right)\mathrm{A}=\mathrm{A}\left({\mathrm{xy}}^{-1}\right) \\ \iff \mathrm{A}={\left({\mathrm{xy}}^{-1}\right)}^{-1}\mathrm{A}\left({\mathrm{xy}}^{-1}\right) \end{gathered}$$

Simplify the equation as:

$$\begin{gathered} \iff \mathrm{A}={\mathrm{yx}}^{-1}{\mathrm{Axy}}^{-1} \\ \iff {\mathrm{y}}^{-1}\mathrm{Ay}={\mathrm{x}}^{-1}\mathrm{Ax} \\ \iff \mathrm{f}\left(\left(\mathrm{H}\cap \mathrm{N}\left(\mathrm{A}\right)\right)\mathrm{x}\right)=\mathrm{f}\left(\left(\mathrm{H}\cap \mathrm{N}\left(\mathrm{A}\right)\right)\mathrm{y}\right) \end{gathered}$$

Thus, f is surjective since if $\mathrm{x}\in \mathrm{H}$implies $\mathrm{f}\left(\left(\mathrm{H}\cap \mathrm{N}\left(\mathrm{A}\right)\right)\mathrm{x}\right)={\mathrm{x}}^{-1}\mathrm{Ax}$.

Since f is bijective, the number of H-conjugates of A equals $\left[\mathrm{H}:\mathrm{H}\cap \mathrm{N}\left(\mathrm{A}\right)\right]$.