How many homomorphisms from $\mathbb Z_4$ to $S_4$?

Question

Let $\varphi: \mathbb{Z_4}\to S_4$ be a homomorphism.

From theory I have:

1. Since $1_{\mathbb{Z_4}}$ is the identity element in $\mathbb{Z_4}$, $\varphi(1_\mathbb{Z_4})=1_{s_4}$.
2. $\varphi(1^{-1})=\varphi(3)=(\varphi(1))^{-1}, \varphi(2^{-1})=\varphi(2)=(\varphi(2))^{-1}$

These don't seem enough to find all homomorphisms. What am I missing?

Answer 1

There are problems with your calculation. You seem to be confusing the additive and multiplicative structure on $\mathbb{Z}_4$. If you say that the inverse of 1 is 3, then this means you consider the additive structure. But then the unit element is 0, not 1.

But I can answer your question. Let $\mathbb{Z}_4$ be the ADDITIVE group of residue classes $\pmod 4$. Then the image of $1$ uniquely determines a homomorphism from $\mathbb{Z}_4$ to another group. This image can be any element whose order divides 4, and exactly those. So your question can be rephrased as follows: how many elements are there in $S_4$ whose order is 1, 2 or 4? There is exactly 1 element of order 1, the identity. There are $\binom{4}{2}= 6$ transpositions and 3 permutations that are products of two disjoint transpositions. So there are 9 elements of order 2. Finally, there are 6 elements of order 4 in $S_4$. So the answer is: there are $1+9+6= 16$ elements of order 1, 2 or 4 in $S_4$, hence 16 homomorphisms from $\mathbb{Z}_4$ into $S_4$.

Answer 2

More generally, let's attempt to enumerate the homomorphisms $f:\mathbb Z_n\to S_m$.

So that a non-trivial homomorphism exists, it is necessary that a non-trivial subgroup of $\mathbb Z_n$ is embedded inside $S_m$.

$1$ is a generator of $\mathbb Z_n$ so knowledge of $f(1)$ fully determines a homomorphism $f$.

$1$ is an element of order $n$ so it must be sent to a permutation $\pi\in S_m$ whose order is a divisor of $n$. Otherwise, the fact that a homomorphism maps identity to identity i.e., $f(0)=\text{id}$ will be contradicted.

Let $\rho (k)$ be the number of elements of order $k$ in $S_m$ then the number of required homomorphisms is: $$\sum\limits_{\text{$k$ is a divisor of $n$}} \rho(k)$$

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In your particular case where $n=4$, we are lucky that $n$ has so few divisors, namely, $1$, $2$ and $4$.

Now count the number of elements of these orders in $S_4$.

$\rho(1)= 1$

$\begin{align}\rho(2) &= |\text{cl}((12))| + |\text{cl}((12)(34))|\\&= \binom{4}{2} +\binom{4}{2} \\&= 12\end{align}$

$\rho(4) = |\text{cl}((1234))| = 3! = 3 $

Note that $\text{cl}(\pi)$ is meant to be read as conjugacy class of $\pi$ in $S_4$.

Anyways, we have $1+12+3=16$ homomorphisms $f:\mathbb Z_4\to S_4$.

Answer 3

You can also think in terms of image of the homomorphism, according to the kernel (first homomorphism theorem).

For $\operatorname{ker}\phi=\{0\}$, you get the three copies of $\Bbb Z_4$ into $S_4$, namely the subgoups of $S_4$ generated by the $4$-cycles: $\langle(1234)\rangle$ (=$\langle(1432)\rangle$), $\langle(1243)\rangle$ (=$\langle(1342)\rangle$) and $\langle(1324)\rangle$ (=$\langle(1423)\rangle$). Each of them can be gotten in two ways, by swapping the two $4$-cycles in the subgroup as images of the generators $1$ and $3$. So, overall $3\times 2=6$ embeddings.

For $\operatorname{ker}\phi=\{0,2\}$, you get $\operatorname{im}\phi\cong C_2$: so, $0$ and $2$ are sent to $()$, while $1$ and $3$ are sent to one same element of order $2$ among the $9$ possible ones (namely: $(12)$, $(13)$, $(14)$, $(23)$, $(24)$, $(34)$, $(12)(34)$, $(13)(24)$, $(14)(23)$).

For $\operatorname{ker}\phi=\Bbb Z_4$, you get the trivial homomorphism: $i\mapsto ()$, for every $i\in \Bbb Z_4$.

Therefore, there are overall $6+9+1=16$ homomorphisms $\phi\colon \Bbb Z_4\to S_4$.