Number of homomorphisms from $S_n$ to $\mathbb Z_n$

Question

How can I find all homomorphisms $f:S_n \rightarrow \mathbb Z_n$ for $n > 4$, where $S_n$ is permutation group on $n$ elements and $\mathbb Z_n$ is group mod $n$?

My guess would be that normal subgroups of $S_n$ for $n>4$ are only $1,A_n,S_n$ and then use property of group homomorphisms that $\ker(f)$ is a normal subgroup.

Answer 1

When $n \geq 2$, the group $S_n$ is generated by transpositions, which all have order $2$ and are all conjugate to each other. So if we have a homomorphism $f : S_n \to A$ where $A$ is an abelian group written additively, then $f$ has the same value on all transpositions, which implies $f$ is completely determined by the single value $f((12))$. When a permutation $\sigma \in S_n$ is a product of $r$ transpositions, we have (i) $2f((12)) = 0$ in $A$ and (ii) $f(\sigma) = rf((12))$.

Conversely, for each $a \in A$ such that $2a = 0$, the mapping $f : S_n \to A$ where $f(\sigma) = ra$ when $\sigma$ is a product of $r$ transpositions, is a homomorphism. The number $r$ is only well-defined modulo $2$, which is okay since $ra$ only depends on $r \bmod 2$. Thus the number of homomorphisms from $S_n$ to an abelian group $A$ equals the number of $a \in A$ such that $2a = 0$.

When $A$ is a finite cyclic group, like $\mathbf Z/(n)$, there are no elements of order $2$ in $A$ when $|A|$ is odd and $1$ element of order $2$ in $A$ when $|A|$ is even. Therefore the number of homomorphisms $f : S_n \to A$ is $1$ when $|A|$ is odd (the trivial homomorphism) and $2$ when $|A|$ is even.

Answer 2

More generally, let us try to enumerate the number of homomorphisms $f:S_m\to \mathbb Z_n$ for $m\geq 5$.

As noted, the kernel of a homomorphism must be a normal subgroup of the domain, there are few possibilities for $\ker(f)$:

1. $\{\text{id}\}$: It can be ruled out because $f$ cannot be one-one; otherwise it would mean that a non-abelian group is embedded in an abelian group (this makes no sense).
2. $ S_m$ : That's just the trivial homomorphism.
3. $A_m$: By first isomorphism theorem, it would mean that $\mathbb Z_2 \cong S_m/A_m\cong f(S_m) \leq \mathbb Z_n$. In other words, a cyclic group of order two is embedded inside $\mathbb Z_n$. You need to count the number of subgroups of order $2$ in $\mathbb Z_n$. If $n$ is even then there is exactly one such subgroup. Otherwise, there is none.

We have enumerated all possible homomorphisms and here's the count:

If $n$ is odd then there's only the trivial one.
If $n$ is even then there are two of them. One of them is trivial. And the other one maps a permutation to $n/2$ or $0$ depending on its parity (odd or even). $\begin{cases}\text{even permutation}\mapsto 0\\ \text{odd permutation} \mapsto\frac n2\end{cases}\tag*{}$

Please comment if I have undercounted anywhere.

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Now let's enumerate the homomorphisms $f:S_4\to \mathbb Z_n$. Our strategy is still the same: Let's find the normal subgroups of $S_4$. Since it's a small group, we can work it by considering conjugacy classes of each cycle type like it's done here.

The possibilities for $\ker(f)$ are:

1. $V_4=\langle (12)(34), (13)(24) \rangle$ (a Klein-four group). By first isomorphism theorem, $S_4/V_4\cong f(S_4)\leq \mathbb Z_n$. Now check that $S_4/V_4\cong S_3$. This is not hard. Hint: Recall that there are only two groups of order $6$ and $(12)V_4$ and $(23)V_4$ don't commute.
   Again, a non-abelian group embedded in an abelian group makes no sense.
2. The other three: trivial, whole group and $A_4$ (which we have already done).

The conclusion remains same for homomorphisms $f:S_4\to \mathbb Z_n$. But this wasn't a fool's errand, you might need this when you try to find homomorphisms from $S_4$ to a non-abelian group.

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Now let's do $f:S_3\to \mathbb Z_n$.

$S_3$ doesn't have any extra normal subgroup besides the three we have dealt with. So our previous conclusions hold yet again!

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For $f:S_2\to \mathbb Z_n$, you can just list them out:

For even $n$, we have the trivial one and this:

$\begin{cases} (12)\mapsto n/2\\ \text{id}\mapsto 0 \end{cases}$

And for $n$ odd, we have only the trivial one.

Thus, our conclusion still holds!

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Conclusion

For $m\geq 2$, there are two homomorphisms $f:S_m\to \mathbb Z_n$ if $n$ is even.
Otherwise, there is only one. $\blacksquare$