I am doing an excercise from Gallian's Contemporary Abstract Algebra that wants me to prove that the group $PSL(2,Z_7)$ is simple. I have reached the point $n_2=7$, $n_3=7$ or $28$ and $n_7=8$.
I found some hints to this problem and one hint to prove $n_3$ cannot be $7$ was that no element of order $7$ can normalize a Sylow $3$ subgroup.
My questions are two:
- Gallian does not use the verb 'to normalize'. I am guessing it means that conjugating the subgroup by this element gives back the subgroup. Correct?
- Can anyone give a hint as to why an element of order 7 does not normalize a Sylow 3 subgroup?
Any help with this will be greatly appreciated. ≈======== Edit. I think I found the answer to question 2 if the answer to 1 is 'yes'. Let x be an element of order 7. If x normalizes the subgroup then it is an element of its normalizer. If $n_3=$ 7 then the order of the normalizer is 24. And since 7 does not divide 24 x cannot be an element of said normalizer and we are done. Is this correct?