Note- $m$ is even.
My attempt: I have shown the kernel of the homomorphisms will contain
elements of the form $aba^{-1}b^{-1}$
Also the elements which can be written as a disjoint cycle of the form $(ab)(cd)$, then it will belong to the kernel of the homomorphism
The elements of the form $(ab)(bc)$ will also belong to the kernel of the homorphism .So all the even permutations will go to an element of order $0$ ( or the kernel)
Now I know, by finding the normal subgroup ,that all the odd permutations will go to an element of order $2$ and all the even permutations will go to the element of order $0$.Now my professor has advised me not to find the kernel first and calculate the homomorphism and to use the idea of $A_n$ since it has not been taught in class.
Thus the only option which is left for me is to show that $\varphi(12) \to a$ an element of order $2$ in $Z_m$, Then $\varphi(13) \to (b)$ where $b$ can be any other element of order $2$ .Then $\varphi(132) \to a+b$ where $(a+b)$ is an element of order $0$ which can hold true if $b=m-a$ like $2$ and $4$ in $Z_6$ .How do I come to a contradiction and show that it is not possible.and all the $2$ cycles will go the same element $a$ ..