$A_4$ is the subgroup of $S_4$ consisting of all even permutations.I know that $A_4$ have 12 elements .

$e = (1),a_1 = (1,2,4),a_2 = (1,4,2),a_3 = (1,2)(3,4),a_4=(1,2,3),a_5=(1,4,3),a_6=(2,3,4),a_7=(1,3)(2,4),a_8=(1,3,2),a_9=(1,3,4),a_{10}=(2,4,3),a_{11}=(1,4)(2,3)$

clearly $e$ is a cyclic subgroup by itself and $\{e,a_1\}$ is also a subgroup because $a_1 = (1,2,4) = (1,4)(1,2)$ and then $a_1^2 = (1,4)(1,2)(1,4)(1,2)$ these are not distinct cycles , do they commute to produce $(1,4)(1,4)(1,2)(1,2) = ee = e$ ? and then how to produce all distinct cyclic groups do we just keep trying ?


How many elements of order $2$ are there? They are the products of two $2$-cycles.There are $\binom{4}{2}$ ways to select the first pair that is switched, and we must divide by two since we are counting twice (when the first pair is the second pair). Thus there are $3$ elements of order $2$ and by that reason $3$ cyclic groups of order $2$.

How many elements of order $3$ are there? These are the $3$-cycles, there are $4$ ways of selecting the element not in the cycle and $2$ ways to arrange them. Thus there are $8$ of these. Since each subgroup uses two of these there are $4$ cyclic subgroups of size $4$.

How many elements are there of order $4$? None, since $4$-cycles are odd permutations.


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