For every divisor $k$ of $60$ $(1 < k < 60)$, $(k = 1,2,3,4,5,6,10,12,15,20,30,60)$
I want to determine whether $A_5$ (Alternating group) has subgroups of order of $k$ of (up to isomorphism If so, describe the structure (by giving the isomorphism type of the group) at least one subgroup of that order.
Also, if not, I want to show why $A_5$ contains no subgroup of that order.
Would like to know if my approach is correct:
My approach: Each non-trivial subgroup of $A_5$ is isomorphic to one of the following:
$$ \mathbb Z/2\mathbb Z, \space \mathbb Z/3\mathbb Z, \space \mathbb Z/2\mathbb Z \times \mathbb Z/2\mathbb Z, \space \mathbb Z/5\mathbb Z, \space S_3, \space D_{10}, \space PSL_2(5)$$
And $A_5$ doesn't have subgroups of order $15, 20, and \space 30$.
Side note: The number of Sylow $2,3,and \space5 $ -subgroups of $A_5$ is $5,10, 6$ respectively.
Order $1:$
The trivial group is isomorphic to any group.
Order $2,3,4,5:$
I'll show that the group $A_5$ has subgroups of the orders $2,3,4,5$ and they're isomorphic to $ \mathbb Z/2\mathbb Z, \space \mathbb Z/3\mathbb Z, \space \mathbb Z/2\mathbb Z \times \mathbb Z/2\mathbb Z, \space \mathbb Z/5\mathbb Z$ respectively.
We know that by Cauchy's theorem, $A_5$ must have elements of order $2$, $2$, and $5$, for e.g. $(1\space 2)(3\space 4), (1\space 2\space 3)$, $(1\space 2\space 3\space 4\space 5))$. Also, these generate cyclic subgroups of orders $2,3,5$. A sylow 2-subgroup of $A_5$ is of order $4$, and since $A_5$ has no elements of order $4$ - because elements of order in $S_4$ are odd permutations, these subgroups must be isomorphic to the Klein 4-group $\mathbb Z/2\mathbb Z \times \mathbb Z/2\mathbb Z$
Order $6:$
If $\pi\in S_5$ is a permutaion of $\{1,2,3\}$, then either $\pi$ or $\pi\circ (4\,5)$ is an even permutation, i.e., $\in A_5$. This allows us to embed $S_3\to A_5$.
Some geometry: $A_5$ is the symmetry group of the dodekahedron. It is possible to select $4$ of its $20$ vertices that make up a regular tetrahedron $T_1$. A rotation of the dodekahedron by $72^\circ$ turns $T_1$ into another such tetrahedron $T_2$. Then the subgroup of $A_5$ that fixes $T_1\cup T_2$ turns out to be of isomorphic to $S_3$.
Without geometry: There is no element of order $6$ in $A_5$, thus if there's a subgroup of order $6$, it would be isomorphic to $S_3 \cong D_6$. Thus, (because we know that a simple of order $60$ has six Sylow 5-subgroups, 24 elements of order $5$, ten Sylow 3-subgroups, and $20$ elements of order $3$) if $Q \in Syl_3(G)$, then $|N_G(Q) : Q| = |Syl_3(G)| = 10$. Thus, $N_G(Q)$ is a subgroup of order $6$ isomorphic to $S_3$.
Order $10:$ The group $A_5$ has subgroups of order $10$, and any subgroup is isomorphic to $D_{10}$.
The number of Sylow $5$-subgroups of $A_5$ is $6$. That means that if $P \in Syl_5(G)$, thus $|G:N_G(P)| = 6$. Thus, the normalizer of a Sylow 5-subgroup is a subgroup of order $10$. Also, we know that a group of order $10$ is either cyclic or isomorphic to $D_{10}$. The group $A_5$ doesn't have elements of order $10$, and so all subgroups of order $10$ in $A_5$ must be isomorphic to $D_{10}$.
Let's be more precise, if we consider the natural definition of $D_{10}$ as the symmetries of a pentagon, we see that it is generated by a rotation and a reflection. The former is a $5$-cycle, while the latter is the product of two $2$-cycles. Both of these generators are even permutations and hence elements of $A_5$.
Order $12:$ I will show that a simple group of order $60$ must have a subgroup of order $12$, and thus it's isomorphic to $A_5$.
$G$ is a simple group of order $60$. If I can show that $G$ has a subgroup of order $12$, then $G \cong A_5$.
Let $P \in Syl_2(G).$ Let's look at two cases (because we know that the number of Sylow 2-subgroups of a simple group of order $60$ is either $5$ or $15$. In the latter case, at least two Sylow 2-subgroups intersect non-trivially.)
Case 1: Let's assume that the number of Sylow 2-subgroups of $G$ is $5$. Thus, $|G: N_G(P)| = 5$, so $N_G(P)$ is a subgroup of $G$ of order $12$, done.
Case 2: Let's assume that the number of Sylow 2-subgroups of $G$ is $15$. Thus, (because we know that the number of Sylow 2-subgroups of a simple group of order $60$ is either $5$ or $15$. In the latter case, at least two Sylow 2-subgroups intersect non-trivially), we will find two distinct Sylow 2-subgroups $R$ and $S$ of order $4$. such that $|R\cap S| = 2$.
Let $x$ be the non-identity element of order $2$ in $R\cap S$. We'll find $|C_G(x)|$. every group of order $4$ is abelian, thus, both $R$ and $S$ are subgroups of $C_G(x)$. Thus, $|C_G(x)|$ is divisble by $4$ and is a divisor of $60$. Thus, $|C_G(x)|$ is $12, 20,$ or $60$. Also, we know that $G$ doesn't have a subgroup of order $20$ (see my proof below for that order). Also, $|C_G(x)| = 60$ means that $x$ is non-trivial element of the center of $G$. But $G$ is simple and non-abelian, and so its center is trivial. We conclude that $|C_G(x)| = 12$, so now we have a subgroup of order 12. done.
Also, the subgroups of order $12$ in $A_5$ are isomorphic to $A_4$. because $A_5$ has a subgroup of order $12$, we also know that $A_4 \le A_5$. Hence, we can say now that $A_5$ does have subgroups of order $12$ isomorphic to $A_4$. ($A_4$ is the even permutation on four objects and hence there are at least five copies of $A_4$ in $A_5$).
Order 15, 20, 30 all together:
The only (proper) divisors of $|A_5|=60=2^2\cdot 3\cdot 5$ greater than $12$ are $15=3\cdot 5, 20=2^2\cdot 5,$ and $30=2\cdot 3 \cdot 5$. Any subgroup of index $2$ is necessarily normal, and groups whose order is the product of three primes are solvable, so since $A_5$ is not solvable so it may not contain a subgroup of order $30$. If $H$ is a subgroup of order $15$, then $H$ is cyclic, but an easy computation shows that in $A_5$ elements of order $5$ and elements of order $3$ do not commute. Therefore it remains to be shown that $A_5$ can't have a subgroup of order $20$.
Now, notice that $A_5$ does have a subgroup $K$ of order $12$ - in particular, $K$ is an isomorphic copy of $A_4$. Suppose that there exists a $J\leqslant A_5$ of order $20$ and notice that $[A_5:J]$ and $[A_5:K]$ are coprime. Then $A_5=JK$ and $[J:J\cap K]=[A_5:K]$. So $|J\cap K|=4$. Thus $J\cap K\in\operatorname{Syl}_2(J)\cap\operatorname{Syl}_2(K)$. We observe that any $H_3\in\operatorname{Syl}_3(K)$ normalizes $J \cap K$, and in particular, $H_3$ acts fixed point freely on $J \cap K$. We see that the Sylow $5$-subgroup $H_5$ of $J$ must normal in $J$, so $J\cap K$ acts nontrivially on $H_5$. It follows that $H_3$ acts faithfully on $H_5$, but this is impossible since $\operatorname{Aut}(H_5)$ has order $4$. This completes the proof.
Order $20:$ (another interesting proof)
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 $20$.
Order $60:$ (with $PSL_2(5)$)
$PSL_2(5)$ can be considered as group of Mobius transformations $$z\mapsto \frac{az+b}{cz+d}, (a,b,c,d\in\mathbb{Z}_5, ad-bc=1)$$ for $z\in \{0,1,2,3,4\}\cup\{\infty\}$, where $\{0,1,2,3,4\}=\mathbb{Z}_5$ and $\infty=1/0=2/0=...=4/0$, and some conventions similar to this.
The group $PSL_2(5)$ acts on these six symbols, whereas we want an action on $5$ symbols to show that it is isomorphic to $A_5$. So how to proceed.
The way I would like to proceedis to do the following:
In $PSL_2(5)$, find an element $A$ of order $2$, $B$ of order $3$ such that product $AB$ has order $5$.
This is quite easier: consider $A(z)=-1/z$. You see that it is of order $2$.
To get element of order $3$, note that if $B^3=I$ in $2\times 2$ matrix form then $B$ satisfies polynomial $t^3-1=(t-1)(t^2+t+1)$; the second factor is irreducible over $Z_5$; so take companion matrix $\begin{bmatrix} 0 & -1 \\ 1 & -1\end{bmatrix}$.
The corresponding Mobius map is $B(z)=(0.z-1)/(1.z-d)=-1/(z-1)$; check- this is of order $3$ (i.e. $B^3(z)=z$)
Finally, consider $AB(z)$; it is $AB(z)=z-1$, great! this is of order $5$ since this permutes $0,1,2,3,4$ in fashion of cycle of order $5$ and leaves $\infty$ fixed (by additive convention with $\infty$: $\infty+a=a+\infty=\infty$).
Thus, we found elements $A,B$ of order $2,3$ respectively such that $AB$ has order $5$. The group $A_5$ has presentation of this form: $$A_5=\langle x_1,x_2: x_!^2=1, x_2^3=1, (x_1x_2)^5=1\rangle.$$ This gives a surjective homomorphism from $PSL_2(5)$ to $A_5$ ($A\mapsto x_1, B\mapsto x_2$).
Finally $PSL_2(5)$ and $A_5$ have same orders; the surjective homomorphism should be isomorphism. Thus, $PSL_2(5) \cong A_5$.