Good evening! I was just wondering if somebody could explain how we are able to make an assumption (that I will mention below) in the following theorem (2.18 in Isaacs' Finite Group theory; it's attributed to Zenkov):
Let A and B be abelian subgroups of a finite group G, and let M be a minimal member of the set {$A \cap B^g$ | $ g \in G$ }. Then $M \subseteq F(G)$.
Note: M is a mininmal member in the sense that no member of the above set is properly contained in M, and F(G) is the fitting subgroup of G.
Isaacs argues that since the above set is unchanged if we replace B by some arbitrary G-conjugate $B^g$, there is no loss to assume that $M = A \cap B$. I was hoping someone could explain why this justification is sufficient to ensure that there is no loss of generality. The rest of the proof is clear to me.
Thank you so much!
Let me try to be a little explicit: Set $$S = \{A\cap B^g : g\in G\},$$ and suppose $M = A\cap B^h$ is a minimal set in $S$. Set $\tilde{B} = B^h$, so $$\tilde{S} = \{A\cap \tilde{B}^g : g\in G\}.$$
Then, you say you understand $S = \tilde{S}$, which means they have the same minimal sets. In particular, $A\cap B^h = A\cap \tilde{B}$ is a minimal set in $\tilde{S}$.
My comment was wrong, you should change $B$ to $B^g$ not $B^{g^{-1}}$.