How many homomorphism are there of the form $S_n \to \mathbb{Z_2},n\geq 5$

Question

How many homomorphism are there of the form $S_n \to \mathbb{Z_2},n\geq 5$?

My solution:

Using first isomorphism theorem each kernel of homomorphism is a normal subgroup in $S_n$.

I know the only normal subgroups of $S_n$ are $1,A_n,S_n$.

Let $q$ be a homomorphism using first isomorphism theorem $\frac{S_n}{\ker q}\cong{\rm Im}(q)\leq \mathbb{Z_2}$.

Case 1: $\ker q= S_n \implies \frac{S_n}{\{S_n\}}\cong{\rm Im}(q)\leq \mathbb{Z_2}\implies{\rm Im}(q)=\{1\}$

Case 2: $\ker q= A_n \implies \frac{S_n}{\{A_n\}}\cong{\rm Im}(q)\leq \mathbb{Z_2}\implies{\rm Im}(q)= \mathbb{Z_2}$.

Case 3: $\ker q=\{1\} \implies \frac{S_n}{1}\cong{\rm Im}(q)\leq \mathbb{Z_2}$, which isn't possible since $|S_n|=n!>|\mathbb{Z_2}|$

There are $2$ homomorphisms: case 1 and case 2.

Is my solution correct?

Thanks!

Answer 1

If you didn't know about normal subgroups of $S_n$, you could've still reached the same conclusion like this:

• Note that in $S_n$ every two $2$-cycles (transpositions) are conjugate to each other. Namely, if $f=(x\,y)$ and $g=(z\,t)$ then you can easily construct $h$ such that $g=h^{-1}fh$. Say, if all $x,y,z,t$ are different, then $h=(z\,x)(t\,y)$. If one pair is the same, say $x=z$, then $h=(t\,y)$. If both pairs are the same, then $h=id_{S_n}$.
• A homomorphism into $\mathbb Z_2$ must map conjugate elements to the same element, as $\mathbb Z_2$ is Abelian. Namely, if $F:S_n\to\mathbb Z_2$ is a homomorphism, and $g=h^{-1}fh$ in $S_n$, then $F(g)=F(h)^{-1}F(f)F(h)=F(h)^{-1}F(h)F(f)=F(f)$.
• That means that all $2$-cycles must map into the same element of $\mathbb Z_2$.

Thus, we have only two cases:

• Every $2$-cycle maps to $0$ and so every other element of $S_n$ maps to $0$ - trivial homomorphism
• Every $2$-cycle maps to $1$, which means that every odd permutation in $S_n$ (a composition of an odd number of $2$-cycles) maps to $1$ (a sum of an odd number of $1$'s) and, similarly, every even permutation maps to $0$.