$A_5$ has no subgroup of order 15 and 20 
Show that $A_5$ has no subgroup of order 15 and 20.  

I have been thinking about this problem for so much time but I'm still clueless. Can anyone tell me how to do this problem? Thanks. 
I looked up online and saw some proofs with simple groups or Sylow Theorem. Can somebody solve this problem without using simple groups or Sylow Theorem? Thanks.
 A: Hint: 
Prove the following:
Claim: If a group $\;G\;$ has a subgroup $\;H\;$ of index $\;n\;$ then it has a normal subgroup contained in $\;H\;$ and of index at most $\;n!\;$
Proof: Hint: make $\;G\;$ act on the left cosets of $\;H\;$, look at the induced homomorphism $\;G\to S_n\;$ and to its kernel...
From the above follows that a simple group cannot have subgroups of index $\;n\;$ if the order of the group doesn't divide $\;n!\;$ .
A: On contrary Suppose there is subgroup of $A_5$ with order $20$ say H .Consider $\alpha \notin H$ and $\alpha$ has order 5.Our Aim is to show that every element of order 5 belong to H as there are 24 elements of order 5 in $A_5$ our assumption will be wrong in this case. Then $A_5$=$H \cup \alpha H \cup \alpha^2H$.
As Now cosider $\alpha^3 \in $ $H$ or $\alpha H$ or $ \alpha^2H$.Which is contradication this is because if $\alpha^3 \in $ $H$ then $(\alpha^3)^2$=$\alpha^6$=$\alpha$ $\in H$ which in not possible .
Similarly $\alpha^3 \in $ $ \alpha^2H$ as this means $\alpha \in H$ which is contradicts 
Similarly $\alpha^3 \in $ $ \alpha H$ as this means $\alpha^2 \in H$ which is contradicts 
Therefore $ \alpha \in H$ this is true for every $ \alpha $ of order 5 hence H contain 24 element of order 5 which is contradiacation with fact that its order is 20
Similar to show about 15 order subgroup not possible 
A: $\langle (1,2,3,4,5), (1,2,3) \rangle = A_5$ notice. If $A_5$ had a subgroup of order 15, then it must (by Cauchy) contain an element of order 5 and an element of  3. But then the order of any group generated by those elements must be of order 15, 30 or 60. It cannot be 30, else then the group would be normal (contradicting simplicity) since it is of index 2. You can show that in fact any three cycle and any 5 cycle must generate the whole group. 
Doing this without Sylow and Simplicity........might be hard. 
