Consider $A_{5}$ group and let $a,b \in A_5 $ .
Find probability of the subgroup generated by a and b. $\langle a, b\rangle$ , be cyclic group .
I think if $a=1$ then we have 60 cyclic subgroup because $A_5$ has 60 elements, also 59 cyclic subgroup for $b=1$ so we have 60+59 cyclic subgroups and probability is $\frac{119}{60×60}$ is this true?
The orders of elements of $A_5$ are
$Order \quad Num\, Elements \\ \\ 1 \quad\quad\quad\quad 1 \\ 2 \quad\quad\quad\quad 15 \\ 3 \quad\quad\quad\quad 20 \\ 5 \quad\quad\quad\quad 24$
Since none of these numbers are composite, $<a,b>$ will be cyclic precisely when one of these elements is a power of the other. This is because if two elements generate a cyclic group and one is not a power of the other, then the cyclic group has composite order (in a group of prime order all elements are powers of any non-identity element, and the identity alone doesn't generate a group of prime order). But $A_5$ has no cyclic subgroups of composite order as it has no elements of composite order.
Since the problem does not state whether $a$ and $b$ are drawn with or without replacement, it's somewhat ambiguous, and it actually makes a rather large difference: $<a,b>$ is substantially more likely to be cyclic if drawn with replacement. However, in usual mathematical usage, one does not assume $a \ne b$ unless explicitly stated, so technically we should answer the problem as though these are drawn with replacement.
Therefore, out of $3600$ (i.e., $60\cdot 60$) cases, we have
$60$ cases of $a=1$ and $b$ any.
$30 = 2\cdot 15$ cases of $o(a)=2$ and $b\in <a>$
$60 = 3 \cdot 20$ cases of $o(a) = 3$ and $b\in <a>$
$120 = 5 \cdot 24$ cases of $o(a) = 5$ and $b\in <a>$
So the probability you are looking for is $270/3600$, or $7.5\%$
Also note that if you want the "draw without replacement answer" you can simply subtract the $60$ cases of $a=b$ from both the numerator and the denominator, to get $210/3540=7/118$, or just a hair over $5.9\%$.