# Exercise about simple group of order 60

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 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”. – 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 $$A_5$$.
• 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. – Beginner Sep 28 '19 at 5:48