Here's exercise 1.41 from Lang's Algebra which I'm trying to figure out.
Let $H$ be a simple group of order $60.$
(a) Show the action of $H$ by conjugation on the set of its Sylow subgroups gives an imbedding $H\rightarrow A_6$.
(b) Show that $H\simeq A_5$.
(c) Show that $A_6$ has an automorphism which is not induced by an inner automorphism of $S_6$.
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I've figured out part (a).
For (b), since $A_6$ is generated by the set of all 3-cycles, can I say $H$ is generated by order 3 elements? Is the subgroup of $H$ generated by order 3 elements normal in $H$?
$H$ has index 6 in $A_6$. What do I need more to conclude that $H\simeq A_5$?
For (c), if every element of $H$ fixed some Sylow 5-subgroup then does $H$ have to be simple?
(I've come across some other posts about similar questions, but I didn't really understand. Please help me with this direction. Thanks.)
If you have proved that $H$ has index 6 in $A_6$, then consider the action of $A_6$ on the left cosets of $H$ by multiplication on the left. This provides an embedding of $A_6$ into $S_6$, and again the image must lie $A_6$. But now the image of $H$ under that embedding is the stabilizer of the coset $H$, so $H$ is isomorphic to a stabilizer of a point in $A_6$, which is $A_5$.
The above embedding actually defines an (outer) automorphism of $A_6$, which maps $H$ onto $A_5$.