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I'm not sure how to continue my proof:

Let $G$ be a simple group of order 60.

Suppose $H \le G $ is of index 5. i want to show that $G $ is isomorphic to $A_5 $

I have homomorphism $f : G \to S_5 $ , $ ker(f) = \{e\}$ so $G $ is isomorphic to $im(f) $.

Now i know that $ im(f) \le S_5$ simple and of order 60.

i also know that $A_5 $ is normal in $ S_5$ so the intersection of $A_5$ and $ im(f)$ is normal in $im(f) $.

$im(f) $ is simple , so this intersection is either $\{e\} $ or $ im(f)$.

I want to show that it is impossible that the intersection is $ \{e\} $ but im not sure how.

EDIT: After some thinking , the answer for this question is not that hard. I added my own answer for it.

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    Why do you want to delete it??? – 2017-01-13
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    So post your own solution as an answer! – 2017-01-13
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    You shouldn't delete your question because you solved it. Like @SteveD mentioned, you're welcome (and even encouraged) to post your own solution as an answer. Perhaps another user will come up with another way of solving this, which may also be of interest to you. And finally, leaving the question and answer helps anyone else with the same question! – 2017-01-13
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    @SteamyRoot i did write my answer. next time, ask the owner of the post if he could edit it back, and dont just modify it. – 2017-01-13
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    @SteveD i did it . – 2017-01-13
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    @Liad All I did was a "rollback" to before your edit that essentially removed the question, which is [the recommended action to take](http://meta.stackexchange.com/questions/209436/self-vandalism-what-is-correct-action) in case of *self-vandalism*. I did not "modify" anything. – 2017-01-13

2 Answers 2

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After some thinking, the answer is quite simple :

The only normal subgroups of $S_5 $ are $\{e\} , A_5 , S_5 $ , and because $im(f) $ has index 2 , it is normal in $S_5 $ thus $im(f) = A_5 $.

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Hint:

Prove that for any $\;H\le S_n\;$ such that $\;H\lneq A_n\;$ , exactly half the elements of $\;H\;$ are even permutations.