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 $S_6$.


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.)

  • $\begingroup$ I don’t understand what you are trying to do in (b). Consider the $5$-Sylow subgroups. Since $H$ is simple, there is more than one. The number of $5$-Sylow subgroups must be a divisor of $60$, and must be congruent to $1$ modulo $5$. The only possibility is that ther are six such subgroups. The action by conjugation on the set gives you an action on a set with six elements, which yields a morphism $H\to S_6$. Now you must show that the map is one-to-one and that the image is actually in $A_6$. You do not know if $H$ is generated by elements of order $3$ yet. All you know is $|H|$ and “simple”. $\endgroup$ – Arturo Magidin Sep 26 '19 at 4:22

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 $A_5$.

  • $\begingroup$ Nice answer; I didn't thought these lines (I thought the proof of isomorphism will be difficult since image of $H$ is transitive group on $6$ letters, whereas the $A_5$ involved in isomorphism is acting on 5 letters; I thought that the isomorphism would be not-obvious or establishing by some internal properties (simple group of order $60$ is $A_5$). However, the answer above gives clear insight for the required isomorphism. $\endgroup$ – Beginner Sep 28 '19 at 5:48

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