Why $A_{5}$ has no subgroup of order 20? Why $A_{5}$ has no subgroup of order 20?
Thanks!
 A: A group of order 20 has a normal Sylow 5-subgroup that is cyclic, $C_5$. We know that the Sylow 2-subgroup of $A_5$ is isomorphic to the Klein 4-group (no 4-cycles in $A_5$).
Assume $G$ is a group of order 20 with the above properties: $C_5\lhd G$, and $V_4\le G$. Then
conjugation in $G$ gives an action of $V_4$ on $C_5$, and hence a homomorphism
$\phi: V_4\rightarrow Aut(C_5)$. We can easily check that the automorphism group of $C_5$ is cyclic of order 4. It is generated by the automorphism $\sigma:x\mapsto x^2$. This implies that $\phi$ cannot be injective, in other words there is an element $g\in V_4$, $g\neq1_G,$ such that
$g$ commutes with all the elements of $C_5$. The group $\langle C_5\cup\{g\}\rangle$ is thus cyclic of order $10$.
But the group $A_5$ has no elements of order ten. Therefore it cannot have a subgroup like $G$ either. Therefore it cannot have any subgroups of order $20$.
A: To the contrary, let $G \leq A_5$ such a group. Then intersection of $G$ with any of the $A_4$s in $A_5$ (there are 5 of them) must be $V_4$. This is due to $|G A_4| = \frac{|G| |A_4|}{|G \cap A_4|}$ and Lagrange's theorem. 
So $G$ contains all double transpositions in $A_5$. But $(12)(34)(34)(15) = (1 5 2)$, a contradiction with Lagrange's theorem.
A: We use the following fact:

Suppose $N \trianglelefteq G$. Then $N$ contains every element which has finite order coprime $[G:N]$.

Suppose there is a subgroup of order $20$ in $A_5$. This gives us an homomorphism from $A_5$ to $S_3$, the kernel $K$ of which has order $10$ or $20$ by the first isomorphism theorem. The index of $K$ is coprime to $5$, which implies that $K$ contains every element of order $5$ since $K$ is normal. But there are $24$ elements of order $5$ in $A_5$, a contradiction.
This approach can be used to give a nice proof of the fact that $A_5$ is simple, which is done in this paper by Gallian. There is a slight improvement to his proof: normal subgroups of order $2$ are ruled out since they are necessarily central and $A_5$ has trivial center.
A: First, yes: it is that simple (ah, how we loathe simple things, uh?) 
Second, of course you've used a very special property of $\,A_5\,$: it is a simpe group! Otherwise how could you deduce $\,A_5\,$ is isomorphic to a subgroup of $\,S_3\,$?? After all, it could be that the kernel of the action you described is big enough as to make this possible, but that kernel is a normal subgroup of $\,A_5\,$, so...
This exercise is more or less middle mathematics undergraduate level (unless, of course, somebody says it is high school's level...we've had some wise guys like these around lately), so you must give more credit to yourself when you reach logically some result. Just be sure every single claim you make you can back it.
A: Your argument is perfectly valid once you accept the simplicity of $A_{5}$. You could also argue this way, and I mention it because it brings out one or two points not mentioned in other answers, though Jyrki's is the closest in spirit. Let us first remark that if $G$ is a finite group, and $P$ is a Sylow $p$-subgroup of $G$, then Sylow's theorem tells us that $[G:N_{G}(P)] \equiv 1$ (mod $p$). From this we may conclude that a group of order $20$ has one Sylow $5$-subgroup, which is necessarily normal.Suppose then that $G = A_{5}$ has a subgroup $H$ of order $20.$ Let $P$ be the unique Sylow $5$-subgroup of $H$, which is also a Sylow $5$-subgroup of $G.$ Now $P \lhd H,$ so that $H \leq N_{G}(P).$ Hence $[G:N_{G}(P)]$ divides $[G:H] =3,$ so that $[G:N_{G}(P)] = 1$ as this is the only divisor of $3$ congruent to $1$ (mod $5$). But this means that $A_{5}$ has a unique Sylow $5$-subgroup, which is certainly not the case, since $\langle (12345) \rangle$ and $\langle (13245) \rangle $ are different Sylow $5$-subgroups of $G.$
It's a useful general fact that if $X$ is any finite group, and $Q$ is a Sylow $q$-subgroup of $X,$ then whenever $M$ is a subgroup of $X$ containing $N_{X}(Q),$ we have 
$[X:M] \equiv 1$ (mod $q$) (just apply Sylow's theorem in $M$ and in $X$).
