Homomorphism from $S_4$ to $S_3$

Question

I was reading Artin's Algebra and stumbled upon this example of a homomorphism $\phi:S_4 \to S_3$. See here and here for the example.

My question is what motivates the partition of the set into pairs of subsets of order 2. The text never bothers to explain the reason behind the move.

In addition, the writer claims that "An element of the symmetric group $S_4$ permutes the four indices, and by doing so it also permutes these three partitions." What does it even mean to permute the partitions? How does permuting the partitions correspond to permuting the indices?

Then I got completely lost when he says $(1\ 2\ 3\ 4)$ acts on the set $\{\Pi_1,\Pi_2,\Pi_3\}$ as $(\Pi_1\ \Pi_3)$. Exactly how are they the same?

Answer 1

To my mind it's cleaner to describe these things geometrically when possible.

$S_4$ naturally acts by rotations and reflections on a tetrahedron, permuting the four vertices. If we want to write down a map $S_4 \to S_3$ in a geometric way, we should find a set of $3$ things in a tetrahedron that rotations and reflections permute.

The clue is that a tetrahedron has $6$ edges: in fact the elements of $S_4$ naturally permute the set of pairs of opposite edges (all of which are visible in the above picture). This corresponds exactly to these partitions into pairs of subsets of size $2$, and hopefully everything else should be much clearer once you have this picture.

When Artin says permuting the indices permutes partitions, he means, for example, that if we have a permutation like $(123)$ that sends $1 \mapsto 2, 2 \mapsto 3, 3 \mapsto 1, 4 \mapsto 4$, it also acts on subsets of $\{ 1, 2, 3, 4 \}$ by permuting each individual element, so for example it sends the subset $\{ 1, 2 \}$ to the subset $\{ 2, 3 \}$, etc. and in this way it also permutes set partitions, e.g. sending the partition $12 | 34$ to the partition $23 | 14$. Again all of these can be visualized in terms of edges of the tetrahedron.

Answer 2

Qiaochu Yuan gave you a nice intuition but let me give you a formal answer (after all the devil is always in details).

Let $\mathbb{N}_m=\{1,\ldots, m\}$ and let $\sigma\in S_m$. If $P=\{A_1,\ldots,A_k\}$ is a partitioning of $\mathbb{N}_m$ then

$$\sigma(P):=\{\sigma(A_1),\ldots, \sigma(A_k)\}$$

is a partitioning of $\mathbb{N}_m$ as well. You may think about $\sigma(P)$ as a "double image". So the idea here is to choose a collection of partitionings $\{P_1,\ldots, P_s\}$ in such a way that $\sigma(P_i)=P_j$ for any $\sigma\in S_m$. Because in this situation $S_m$ acts on $\{P_1,\ldots, P_s\}$.

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In this concrete case we have $m=4$ and we deal with special partitionings: such that every element of the partitioning is a subset of size $2$. So how many such partitionings are there? Well, since we deal with a set of size $4$ then every such partitioning is uniquely determined by chosing a single subset of size $2$. And there are $6$ such subsets but each subset is paired with its complement which actually gives our three $\Pi_1,\Pi_2,\Pi_3$ partitionings. These are all there are under these constraints.

And since every permutation maps a subset of size $2$ to a subset of size $2$ and these are all possible partitionings into subsets of size $2$ then every permutation "double maps" each $\Pi_i$ to some $\Pi_j$. In other words $S_4$ acts on $\{\Pi_1,\Pi_2,\Pi_3\}$.

This induces a map $f:S_4\to S(\{\Pi_1,\Pi_2,\Pi_3\})$. Now when he says that $(1,2,3,4)$ acts on $\{\Pi_1,\Pi_2,\Pi_3\}$ as $(\Pi_1,\Pi_3)$ then it simply means that $f((1,2,3,4))=(\Pi_1,\Pi_3)$ or in other words that $(1,2,3,4)$ "double maps" $\Pi_1$ to $\Pi_3$. Note that the notation $(\Pi_1, \Pi_3)$ denotes the permutation $\tau:\{\Pi_1,\Pi_2,\Pi_3\}\to \{\Pi_1,\Pi_2,\Pi_3\}$ such that $\tau(\Pi_1)=\Pi_3$.

Answer 3

There is a lot going on here, and it illustrates some important topics. Rather than repeat what others have said, I'll note a few things. The geometric point of view is pertinent, and groups have a key role in describing the symmetries of things. I will use $V$ for the special subgroup of order $4$.

First I would note that there is a specific phenomenon here unique to $S_4$ - it is the only symmetric group $S_n$ which has a proper normal subgroup other than $A_n$, and $A_4$ is the only alternating group which is not simple. The existence of this special subgroup $V$ is connected with the fact that a quartic equation is "solvable by radicals" - by taking square roots and cube roots and fourth roots. $V$ is the smallest group which is not cyclic. ($S_3$ is the smallest which is not abelian).

Note also that the elements $(1, 2); (3, 4); (1, 2)(3, 4)$ of $S_4$ together with the identity form a subgroup with the same structure as $V$ - the non-identity elements all have order $2$, and it is somewhat easier to see at first glance what is going on. However the elements here have different cycle-types. The significance of $V$ is that all the non-identity elements have the same cycle type.

Now if you have a collection of all the elements of the same cycle-type in a symmetric group, the group acts on those elements by conjugation. The conjugate of any element is an element with the same cycle-type. You may not yet have studied material on group actions, but a group action on any set with $n$ elements involves a homomorphism from the group to some subgroup of $S_n$. Here, seeing precisely three elements of the same cycle-type, we know there is a homomorphism from $S_4$ to a subgroup of $S_3$ induced by this action - the action of conjugation permutes the three non-identity elements of $V$.

It so happens that $S_4$ has order $24$ and the only elements of $S_4$ which fix $V$ pointwise by conjugation are the elements of $V$. $V$ has order $4$ and is the kernel of the homomorphism. The order of the image is $24/4=6$ and must therefore be the whole of $S_3$.

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To go back to the question and how permuting the indices permutes the partitions, any subgroup of $S_4$ rearranges the underlying symbols - let's call them $a, b, c, d$ because numbers sometimes get confusing here.

Think about what happens to $(a, b)(c, d)$ when things are rearranged. If, for example $a$ and $b$ go to $b$ and $d$, then $c$ and $d$ must go to $a$ and $c$ in some order, and you get $(b, d)(a, c)$ of the same type.

The geometric point of view makes this obvious in the pairs of opposite sides of a tetrahedron.

How did this first get noticed? - Well probably many different ways by many different people. I mentioned the solution of equations by radicals because that is one of the reasons group theory got off the ground - with people studying the permutations of roots of polynomial equations.

The significance of $V$ is both that it is small and exceptional (and therefore interesting) and also that it is a first illustration of more general ideas (eg group actions, normal subgroups)