Question on finding more than one group homomorphisms from one group to another

Question

Question:

How many group homomorphisms from $(\mathbb Z/3, +)$ to $S_4$ are there?

My attempts:

For the group of $\mathbb Z/3$ with addition as the operation, I know that it refers to the group of modulo 3.

And for $S_4$, I think it refers to the symmetry group of the all the permutations of 4 elements, which is a group consisting of 24 elements.

And I know that for the group $(\mathbb Z/3, +)$ to be a homomorphism to $S_4$, I need to check if $f(ab) = f(a)f(b)$, for f being the mapping from group $(\mathbb Z/3, +)$ to $S_4$ and $a, b $ are any elements in $(\mathbb Z/3, +). $

My Question 1: As I am not sure on the operation in $S_4$, I am quite stuck at the checking here.

My Question 2: If $f(ab) = f(a)f(b)$ is true, it stands for there is only ONE homomorphism exist from $(\mathbb Z/3, +)$ to $S_4$, right? But I think the question gave hints that there is more than ONE homomorphism from $(\mathbb Z/3, +)$ to $S_4$. So how to find out the other group homomorphisms from $(\mathbb Z/3, +)$ to $S_4$?

Please assist, thank you so much!

Answer 1

This is based on group actions.

A homomorphism $\varphi\colon \Bbb Z_3 \to S_4$ is equivalent to an action $\mathcal{A}\colon \Bbb Z_3\times X\to X$, where $X:=\{1,2,3,4\}$. Now, the stabilizers are subgroups of the group which acts, so for every $i\in X$, either $\operatorname{Stab}(i)=\{0\}$ or $\operatorname{Stab}(i)=\Bbb Z_3$; accordingly, an orbit's size is either $3$ or $1$, respectively (Orbit-Stabilizer theorem). But the set of the orbits is a partition of $X$, and hence the only allowed orbit equations are $4=1+1+1+1$ and $4=1+3$. The former corresponds to the trivial homomorphism (if all the stabilizers are equal to $\Bbb Z_3$, then the kernel of the homomorphism is $\bigcap_{i=1}^4\operatorname{Stab}(i)=\Bbb Z_3$, and all the group elements are mapped to $()_{S_4}$); the latter corresponds to four embeddings, in fact: a) three out of the four stabilizers are trivial; b) the singleton orbit can be any among $O(i), i=1,2,3,4$.

Answer 2

1. The group operation for $S_4$ is composition of functions. Elements of $S_4$ are just bijective functions $\{1,2,3,4\} \longrightarrow \{1,2,3,4\}$, so this is sensible.

2. The homomorphism property does not imply that there is only one homomorphism from $\mathbb Z/3$ to $S_4$. In fact, there are many. Here's how I would go about finding them.

So the elements of $\mathbb Z/3$ are the equivalence classes $\{0, 1, 2\}$. Now, in this group we of course have $1+1 = 2$ and $1+1+1 = 0$. So supposing we had a group homomorphism $f: \mathbb Z/3 \longrightarrow S_4$ then by the definition of a group homomorphism, $f(2) = f(1+1)=f(1)^2$ and $f(0) = f(1+1+1)=f(1)^3$. In other words, any group homomorphism is completely determined by what $f(1)$ is, as the values of $f(2)$ and $f(0)$ are implied by the value of $f(1)$.

So we've determined that to define a group homomorphism $f: \mathbb Z/3 \longrightarrow S_4$ we need only specify the value of $f(1)$, and the rest follows. So does that mean that there are 24 such group homomorphisms - one per element of $S_4$? No! For instance, let $f(1) = (12)$ (here I am using cycle notation). Then by what we determined above, $f(2)=(12)^2=id$ and $f(0)=(12)^3=(12)\neq id$. And there's our problem, every group homomorphism has to send the identity element of the domain to the identity element of the codomain (I'll leave this as an exercise). What this tells us is that our map $f$ must satisfy $id = f(0) = f(1+1+1)=f(1)^3$. Furthermore, it turns out that for every $\sigma \in S_4$ such that $\sigma^3 = id$, $f(1) = \sigma$ defines a group homomorphism from $\mathbb Z/3 \longrightarrow S_4$. There is more abstract and general reasoning you can apply to prove that, but these groups are pretty small so you could instead find all $\sigma \in S_4$ and check manually that $f(1) = \sigma$ yields a well defined group homomorphism. I'll give you a hint: there are exactly 9 such elements. Thus, there are 9 such group homomorphisms.

So this method I'm using is a special case of something more general, which was hinted in the comments. I first said that where 1 is sent determines where the rest of the group is sent. This means that $1$ "generates" $\mathbb Z/3$. Another examples is $\mathbb Z/3 \times \mathbb Z/3$, which is generated by the set $\{(1, 0), (0, 1)\}$. I then noted that not every choice of where 1, the generator, is sent is valid. This is because $1 \in \mathbb Z/3$ satisfies the "relation" $1+1+1=0$. Thus, whatever it is mapped to must satisfy the same relation on the other side: $f(1)*f(1)*f(1)=id$.

I can't go into much more detail than that without knowing your level of knowledge, but to learn more about this very powerful general method, you would want to learn about free groups and quotients.

Answer 3

The isomorphism theorem gives a handy framework with which to approach this sort of problem. We are searching for homomorphisms $\phi: \mathbb{Z}_3 \to S_4$ wherein:

$$\mathbb{Z}_3/\ker(\phi) \cong \text{Im}(\phi)$$

We know that the kernel of a group homomorphism $G \to H$ is a normal subgroup of $G$; in particular, the size of the kernel must divide the order of $G$. This forces the kernel of a homomorphism $\mathbb{Z}_3 \to S_4$ to either be trivial or contain all of $\mathbb{Z}_3$.

If the kernel is all of $\mathbb{Z}_3$, this corresponds to the trivial homomorphism (which always exists between any two groups), where $\phi(g) = \{ \text{id} \}$ for all $g \in \mathbb{Z}_3$. If the kernel is trivial, then $\phi$ is injective, and we have $\mathbb{Z}_3 \cong \text{Im}(\phi)$, where $\text{Im}(\phi)$ is necessarily a subgroup of $S_4$. Hence, we need to figure out how many distinct copies of $\mathbb{Z}_3$ exist as subgroups of $S_4$.

This can be tackled in a number of ways. The most straightforward approach would be to just count the number of unique $3$-cycles in $S_4$ while bearing in mind that each distinct copy of $\mathbb{Z}_3$ contains a pair of $3$-cycles (plus the identity).

Another approach uses the fact that the rotational symmetry group of a cube is isomorphic to $S_4$. This is very useful to know even outside of the context of this problem (which is why I bring it up here) because it provides a handy visual reference for this permutation group, rendering the group potentially easier to explore than when getting lost in a mess of cycle notation. Anyway, suppose, for example, that you take a rotational axis to be through the middle of any $2$ opposing faces of a cube. Then a $90$-degree rotation about this axis has order $4$, and per the isomorphism, each such rotational axis corresponds$^\ddagger$ to a distinct copy of $\mathbb{Z}_4$ existing as a subgroup of $S_4$. Likewise, copies of $\mathbb{Z}_3$ appearing as subgroups of $S_4$ correspond to the rotational axes of a cube about which there exist rotations of order $3$. Well, how many such axes are there?

The total number of nontrivial homomorphisms $\mathbb{Z}_3 \to S_4$ should be double the number of copies of $\mathbb{Z}_3$ that appear as subgroups of $S_4$. To wit, we choose one the copies of $\mathbb{Z}_3$ inside of $S_4$ to which we can map $\mathbb{Z}_3$ injectively, and there are two ways to define the mapping because $\mathbb{Z}_3$ has two elements that can serve as generators.

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$\underline{\textbf{Footnotes}}:$

$^\dagger$Any homomorphism $\phi:G \to H$ is determined by its action on a generating set for $G$. Because a single element can generate $\mathbb{Z}_n$, determining where this element goes determines any homomorphism out of $\mathbb{Z}_n$. This is because any $a \in \mathbb{Z}_n$ can be written as $g^k$ for some $k$, and we have $\phi(g^k) = \phi(g)^k$. Therefore, this enumerate-copies-of-$\mathbb{Z}_3$ approach is equivalent to figuring out how many elements of order $3$ in $S_4$ there are to which we can send the generator, as per J.W. Tanner's comment.

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$^\ddagger$ Note that, as is discussed in the hyperlink, you can figure out which permutations correspond to which rotations by labeling the $4$ diagonals of the cube.