Even and Odd Maps in Group Homomorphism

Question

I'm curious about a particular property in a group homomorphism. Suppose that $\phi:S_{6}\rightarrow\mathbb{Z}_{2}$ is any map, not necessarily a group homomorphism, with $(S_{6},\circ)$ and $(\mathbb{Z},+)$ are groups and $S_{6}=\{\sigma_{k}\mid k\in[1,6!]\cap\mathbb{Z}\}$ and $\mathbb{Z}_{2}=\{[0]_{2},[1]_{2}\}$. Then, I want to find a particular $\phi$ such that $\phi$ is a group homomorphism. In some sense, it is straightforward that I can assume a map $\phi$ such that each $\sigma_{2k}\mapsto[0]_{2}$ and each $\sigma_{2k+1}\mapsto[1]_{2}$. For such $\phi$, how can I proof that it is indeed a group homomorphism?

Answer 1

The question having been clarified, I can write a more appropriate answer. I'll leave my no-longer-relevant answer at the bottom, so the comments will still make some sense.

There are two, and only two, homomorphisms from $S_6$ to ${\bf Z}_2$. One of them takes every element of $S_6$ to zero, but that's not the one you want. The other one must take to zero a subgroup of $S_6$ of index two in $S_6$ (and take all the other elements of $S_6$ to the element $1$ in ${\bf Z}_2$).

Now, $S_6$ has only one subgroup of index two (that is, one subgroup containing exactly half the elements of $S_6$), and that is the alternating group $A_6$. $A_6$ consists of the "even" elements of $S_6$, that is, the elements that can be written as products of an even number of transpositions. This is easily seen to be a subgroup, as it contains the identity, and the product of two products of an even number of transpositions is a product of an even number of transpositions, likewise for the inverse.

To be specific, the elements of $A_6$ are the $3$-cycles, of which there are $40$; the $5$-cycles, of which there are $144$; the products of two disjoint $3$-cycles, of which there are $40$; the products of two disjoint transpositions, of which there are $45$; the products of a transposition and a disjoint $4$-cycle, of which there are $90$; and the identity element; all told, $40+144+40+45+90+1=360$.

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No, the symmetric group on six letters is not isomorphic to the cyclic group of order two, no matter what map $\phi$ you come up with. For one thing, in order for two groups to be isomorphic, they must have the same number of elements, which these two groups patently do not.