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$S_7$ does not contain a subgroup of order 15

This is an example of converse of Lagrange's Theorem not working.

I want to know how we can prove this.

Do you know which idea we should use?

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up vote 4 down vote accepted

A group of order $pq$ with $p=3$ not dividing $q-1 = 4$ is commutative and therefore cyclic of order 15. Using the cycle decomposition of permutations, this cannot be done with actions on 7 elements. In $S_8$ the product of a disjoint 3-cycle and 5-cycle is an element of order 15.

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