(I compose the permutations left-to-right)
By way of induction on $n$, suppose for all $l < n$ that if $1 < k < l$ then $\langle (1,\ldots,k),\,(1,\ldots,l)\rangle$ contains a $3$-cycle. There is nothing to show for the base case $l=3$. Consider that $$(1,\ldots,n)(1,\ldots,m)^{-1} = (m,m+1,\ldots,n). $$ and $$(1,\ldots,n-m+1) = (1,\ldots,n)^{m-1}(m,m+1,\ldots,n)(1,\ldots,n)^{1-m}, $$ so that $(1,\ldots,n-m+1)\in G$. If $m \neq n-m+1$, then either $1 <m < n-m+1 < n$ or $1 < n-m+1 < m < n$. In either of these cases, the induction hypothesis provides that $\langle(1,2,\ldots,m) (1,\ldots,n-m+1)\rangle$ contains a $3$-cycle, which then must also be contained in $G$. We now only have the case where $n = 2m-1$.
I can proceed no further.
Notice $(1,2, ... , n)(1,2, ..., m)^{-1}(1,2, ..., n)^{-1}(1, 2, ..., m) =(1nm)$, left to right.