Making an exhaustive list of the elements of $A_5$ I know that the elements of the alternating group $A_5$ are defined by the even permutations on 5 elements.
So I understand that $Id$ and $(ab)(cd)$ are elements of $A_5$ but I don't know how to get an exhaustive list.
 A: Let $p$ be an element of $S_5$ other than the identity.

First suppose $p$ is a cycle.

Then $p$ is an even permutation if and only if the cycle length of $p$ is odd. Thus, $p$ is either a $3$-cycle (a b c) or a $5$-cycle (a b c d e).

$\;\;\;{\small{\bullet}}\;\;$Number of $3$-cycles:
$\;{\displaystyle{(2){\small{\binom{5}{3}}} = 20}}$

$\;\;\;{\small{\bullet}}\;\;$Number of $5$-cycles:
$\;{\displaystyle{4! = 24}}$

Next suppose $p$ is a product of nontrivial disjoint cycles.

Since there only $5$ elements in $\{1,2,3,4,5\}$, there can't be more than two nontrivial disjoint cycles in the disjoint cycle representation of $p$.

Since the product of a $3$-cycle and a disjoint $2$-cycle would be yield an odd permutation, it follows that for this case, $p$ is an even permutation if and only if $p$ = (a b)(c d), the product of two disjoint $2$-cycles.

$\;\;\;{\small{\bullet}}\;\;$Number of products of two disjoint $2$-cycles:
$\;{\displaystyle{\frac{{\large{\binom{5}{2}\binom{3}{2}}}}{2}=15}}$

Finally, we have the identity element. 

Adding up the counts to check, we get
$$20 + 24 + 15 + 1 = 60$$
as expected.

To recap, an element $p$ other than the identity is in $A_5$ if and only if $p$ has one of the disjoint cycle representations
$$(\text{a}\;\;\text{b}\;\;\text{c})$$
$$(\text{a}\;\;\text{b}\;\;\text{c}\;\;\text{d}\;\;\text{e})$$
$$(\text{a}\;\;\text{b})\;(\text{c}\;\;\text{d})$$
An explanation of the counts . . .

$\small{\bullet}\;\,$Counting the $3$-cycles:


*
The number of $3$-element subsets $\{\text{a},\text{b},\text{c}\}$ of $S_5$ is ${\large{\binom{5}{3}}}$. Once the elements $\text{a},\text{b},\text{c}$ are chosen, two distinct $3$-cycles can be formed, namely: $(\text{a}\;\text{b}\;\text{c})$ and $(\text{a}\;\text{c}\;\text{b})$, hence, the count is 
$$2{\small{\binom{5}{3}}} = (2)(10) = 20$$
 

$\small{\bullet}\;\,$Counting the $5$-cycles:


*
Each $5$-cycle $(\text{a}\;\text{b}\;\text{c}\;\text{d}\;\text{e})$ can be represented in $5$ ways. For example
$$(1\;2\;3\;4\;5) = (2\;3\;4\;5\;1) = (3\;4\;5\;1\;2) = (4\;5\;1\;2\;3) = (5\;1\;2\;3\;4)$$
Thus, the number of $5$-cycles is 
$$\frac{5!}{5}=4!=24$$


$\small{\bullet}\;\,$Counting the products of two disjoint $2$-cycles:


*
There are ${\large{\binom{5}{2}}}$ of choosing the first $2$-cycle, followed by ${\large{\binom{3}{2}}}$ of choosing the second one. But the same two $2$-cycles could have been chosen with the second one first, and the first one second. Hence, for this case, the correct count is
$$\frac
{{\large{\binom{5}{2}\binom{3}{2}}}}
{2!}
=\frac{(10)(3)}{2}=15$$

