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Let $S_8$ be the symmetric group of permutations of an $8-$element set and let $k$ be the number of its abelian subgroups of order $16$. Prove that $k ≥ 1050.$

I think the subgroups of order 16 are generated by two disjoint cycles of order $4$. The number of such cycles is, I think, $\frac{8\cdot7\cdot6\cdot5}{4 }\frac{3!}{2}=1260$. Thus, is the answer proved. Note that I think the disjoint cycles generate an abelian group. Any ideas. Thanks beforehand.

P.S.:This is problem U396 in Mathematical Reflections

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    what about four disjoint transpositions? – 2017-01-09
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    @JorgeFernándezHidalgo thanks for pointing out, but is my reasoning partially correct? – 2017-01-09
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    i think you need to divide that by $2$ one more time, I think it should be $\binom{8}{4}\times 3!^2 /2$ – 2017-01-09
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    @JorgeFernándezHidalgo how come? the number of cycles of order $k=n\ldots(n-k+1)/k$, isnt it? – 2017-01-09
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    yes, but you want to select unordered pairs of them, so you are counting each pair twice – 2017-01-09
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    @JorgeFernándezHidalgo ok thanks, modified the post. – 2017-01-09
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    There is another possibility: a four cycle with two disjoint transpositions. <(1234),(56),(78)> – 2017-01-09
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    @gobucksmath yes, I agree. But, isnt even this partial enumeration sufficient for the problem? – 2017-01-09
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    Oh, and you also forgot to divide by $9$, because every such subgroup is generated by $9$ different pairs of elements. – 2017-01-09
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    @JorgeFernándezHidalgo could you please elaborate? – 2017-01-09
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    sure, notice that $\langle (1,2,3,4), (5,6,7,8)\rangle = \langle (1,3,2,4),(5,6,7,8)\rangle $. – 2017-01-09
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    @JorgeFernándezHidalgo but that is included in the formula for disjoint cycles, isnt it? – 2017-01-09
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    yes, but given a specific subgroup there can be many ways to select the cycles. – 2017-01-09
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    @JorgeFernándezHidalgo how do you get $9$ different pairs? – 2017-01-09
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    Is this from an ongoing contest? See comment [there](http://math.stackexchange.com/q/2092988/). – 2017-01-11

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