Is there any neat way to show $\phi$ is a homomorphism?

Question

In Michael Artin's Algebra (chapter 2, page 50, example 2.5.13) the author illustrates a homomorphism from $S_4$ (all permutations of indices $(1,2,3,4)$) to $S_3$ (all permutations of indices $(1,2,3)$), as is shown in the picture below:

I understand how the map sends an element from $S_4$ into $S_3$, but I just don't understand why this is a homomorphism. The author remarks that

If $p,q$ are elements of $S_4$, the product $pq$ is the composed permutation $p\circ q$, and the action of $pq$ on the set $\{\Pi_1,\Pi_2,\Pi_3\}$ is the composition of the actions of $q$ and $p$. Therefore $\phi(pq)=\phi(p)\phi(q)$.

But it isn't clear to me how $\phi(p),\phi(q)$ can be so "composed"... It might not be hard to imagine, but how to sketch a rigorous proof?

Of course there are finite cases so we can always do it case by case, but I'm wondering if there is a more elegant way to do this?

Answer 1

Any element of $S_4$ sends any partition of $\{1,2,3,4\}$ to a similar partition of $\{1,2,3,4\}$.

For example, any $\sigma \in S_4$ sends a $3+1$ partition of $4$ to a $3+1$ partition of $4$: if $\sigma = (1\ 3\ 4)$, for example, $\sigma$ sends the $3+1$ partition $\{\{3\},\{1,2,4\}\}$ to the partition: $\{\{\sigma(3)\},\{\sigma(1),\sigma(2),\sigma(4)\}\}= \{\{4\},\{1,2,3\}\}$.

Since the mappings in $S_4$ are bijective, we induce a bijective map on the set of all partitions of any given type, that is: an action on the set $P$ of these partition-types, which is by defintion a homomorphism $S_4 \to P$.

The possible partition-types of $4$ are these:

$4$ = $\{\{a,b,c,d\}\}$-there are only one of these, and this induces the trivial homomorphism $S_4 \to S_1$.

$3+1 = \{\{a\},\{b,c,d\}\}$-there are $4$ of these and this induces an isomorphism $S_4 \to S_4$.

$2+2 = \{\{a,b\},\{c,d\}\}$-there are $3$ of these, and this induces a homomorphism $S_4 \to S_3$.

$2+1+1 = \{\{a,b\},\{c\},\{d\}\}$-there are $6$ of these and this induces a homomorphism $S_4 \to S_6$.

$1+1+1+1 = \{\{a\},\{b\},\{c\},\{d\}\}$-there is just one of these, and it also induces the trivial homomorphism.

A similar technique to this, is used to find outer automorphisms of $S_6$.

EDIT: I was going to add this, this morning, but was rushed for time. It's clear we have an action on $S_3$ by letting $S_4$ induce an action on the set of $2+2$ partitions of $\{1,2,3,4\}$, but all this guarantees is we have a homomorphism into $S_3$. Let's see that this homomorphism is indeed surjective.

Denote $\{\{1,2\},\{3,4\}\} = a, \{\{1,3\},\{2,4\}\} = b, \{\{1,4\},\{2,3\}\} = c$.

Let's examine the mapping of $\{a,b,c\}$ to itself induced by $\sigma = (1\ 2)$.

Clearly, $(1\ 2)$ sends $a \to a$, and $b$ gets sent to:

$\{\{\sigma(1),\sigma(3)\},\{\sigma(2),\sigma(4)\}\} = \{\{2,3\},\{1,4\}\} = c.$

Similarly, $(1\ 2)$ sends $c$ to:

$\{\{\sigma(1),\sigma(4)\},\{\sigma(2),\sigma(3)\}\} = \{\{2,4\},\{1,3\}\} = b$.

Thus $(1\ 2)$ induces the $2$-cycle $(b\ c)$.

One can verify in the same way that $(2\ 3)$ induces the $2$-cycle $(a\ b)$.

Since $S_3$ is generated by any $2$ distinct transpositions, this shows our homomorphism is onto.

Order (via Lagrange and the fundamental isomorphism theorem) arguments show the kernel has order $24/6 = 4$.

One can verify directly that:

$\tau = (1\ 2)(3\ 4)$ acting on $\{a,b,c\}$ clearly sends $a \to a$, and sends:

$b \to \{\{\tau(1),\tau(3)\},\{\tau(2),\tau(4)\}\} = \{\{2,4\},\{1,3\}\} = b$

(sets don't care which order their elements are listed in).

Thus $\tau$ must send $c \to c$, that is, $\tau$ induces the identity map on $\{a,b,c\}$-so $\tau$ is in the kernel of our homomorphism.

Since the kernel (being a kernel) is a normal subgroup of $S_4$, it thus contains all conjugates of $\tau$, that is, all $2,2$-cycles (of which there are $3$). The identity map "fills it up", we have $4$ elements, and can have no more (although you can verify in the same manner that the other $2,2$-cycles also induce the identity map on $\{a,b,c\}$.

Answer 2

$S_n$ is generated by $q=(1 2)$ and $p=(1 2\cdots n)$. In other words, every element can be written as a string of these two. Thus you only need check the two-and threefold compositions of these elements satisfy the equation you name:

$\phi(p^3)=\phi(p)^3$

$\phi(p^2q)=\phi(p^2)\phi(q)$

$\phi(qp^2)=\phi(q)\phi(p^2)$

$\phi(p^2)=\phi(p)^2$

$\phi(pq)=\phi(p)\phi(q)$

$\phi(qp)=\phi(q)\phi(p)$.

Since any cycle in $S_n$ can be written (non-uniquely) as a word in $p$ and $q$, the result will follow from induction on the length of the word and the above. Note that if a cycle can be written as a word in $p$ and $q$ with an even number of letters, we only need refer to the last three equations.

Answer 3

The point is that there are exactly $3$ possible pairings of $4$ objects. Every permutation of the $4$ objects will correspond to mapping one pairing to another, and so is essentially a permutation of the pairings. Composition of permutations will then also correspond to the composition of the corresponding mappings, and so it is a homomorphism from $S_4$ to $S_3$.

Answer 4

Any permutation $\pi\in{\cal S}_4\bigl([4]\bigr)$ acts on the elements of $[4]$, and there is no simple formula describing the resulting action of $\pi$ on the set $P$ of pairings of $[4]$. Therefore I shall realize the pairings as edge colorings of the complete graph $K_4$ on the vertex set $[4]$, as follows:

Identifying an element of $P$, i.e., a pairing of $[4]$, is the same as coloring two disjoint edges of $K_4$ red. Denote the set of possible such colorings by $P'$. Given any permutation $\pi:\>[4]\to[4]$, this permutation will automatically induce a permutation $\phi_\pi\in{\cal S}_3(P')$, and it is then obvious that $$\iota:\quad {\cal S}_4\bigl([4]\bigr)\to{\cal S}_3(P'),\qquad \pi\to\phi_\pi$$ is a homomorphism. The question whether $\iota$ is surjective has not been addressed.