1
$\begingroup$

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.)

$\endgroup$
  • $\begingroup$ 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
1
$\begingroup$

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 $A_5$.

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.