Show that S6 has at least 60 subgroups of order 4 Show that $S_6$ has at least $60$ subgroups of order $4$. [Hint: Consider cyclic
subgroups generated by a 4-cycle (such as $\langle(1234)\rangle$) or by the product of
a 4-cycle and a disjoint transposition (such as $\langle(1234)(56))\rangle$; also look at
noncyclic subgroups, such as ${(1), ( 12), (34), (12)(34)}$.]
For my work so far, I have found that the cyclic groups of order 4 are
$\langle(1234)\rangle, \langle(2345)\rangle,...,\langle(6123)\rangle$ which are $6$ of them
and for disjoint transposition of order 4 I have 
$\langle(1234)(56)\rangle, .... ,\langle(6123)(45)\rangle$ which are also 6 of them
I not sure what the last hint is leading and I don't know how to proceed from here. 
Side note: this is a question from the section related to Sylow's Theorems and Cauchy's Theorem. But I don't think they are helpful for this question. I feel it is just related to symmetric and alternating groups.
 A: For merely the cyclic part, consider elements of the form $(abcd) \in S_4$. How many of them are there? Indeed, the answer seems to be just $6 \times 5 \times 4 \times 3 = 360$ by choosing $a,b,c,d$ in that order. However, noting from the comment that $(abcd) = (bcda) = (cdab) = (dabc)$ ( in the cycle representation, for example $(1234) = (2341)$ ) tells you that we must divide by $4$ to avoid repetition. This leads to $90$ cycles.
However, when we are looking at the generated cyclic group, indeed note that $(abcd)$ and $(adcb)$ generate the same group (because if $x$ is of order $4$ then $x^3 = x^{-1}$ is also of order $4$ and hence generates the same group. However, $x^2$ is not, so we will have to be careful there).
Also, if two elements generate the same subgroup of order $4$ then either they are the same or inverses. Therefore, the above analysis gives $45$ distinct groups of order $4$.

Now, for the other part, we consider groups generated by $(abcd)(ef)$ which would be of order $4$ but would not coincide with any of the previous groups since there is always an element which does not have any fixed points here.
Choosing $(abcd)(ef)$ happens again in $90$ ways, since if we choose $(abcd)$ then $(ef)$ gets fixed for us. Once again, going to the cyclic subgroup , we can pair $(abcd)(ef)$ with its inverse which is $(adcb)(ef)$ for each element of this kind, once again resulting in $45$ distinct groups of order $4$.
Totalling the above gives $90$ distinct cyclic groups of order $4$, more than what is required to answer the question.

Apart from this if we choose to look at groups generated by two transpositions $(ab) \neq (cd)$, which are distinct from the previous ones since every element here has order at most two , then $(ab)$ has $15$ choices and $(cd)$ has $6$ choices, after which you remove their order  of picking to get $45$ further choices.  This leads to at least $135$ subgroups of order $4$. You can try to check if some others are there, my guess would be no.
