This was in my textbook. I don't quite understand it and would like some clarification on it:

Let $\phi: S_{4} \rightarrow S_{3}$ be a homomorphism between symmetry groups. There are three ways to partition the set of four indices $\{1,2,3,4\}$ into pairs of subsets of order two, namely

$ \Pi_{1}: \{1,2\} \cup \{3,4\}$

$ \Pi_{2}: \{1,3\} \cup \{2,4\}$

$ \Pi_{3}: \{1,4\} \cup \{2,3\}$.

An element of the symmetric group $S_{4}$ permutes the four indices, and by doing so it also permutes these three partitions. This defines the map $\phi$ from $S_{4}$ to the group of permutations of the set {$\Pi_{1}$,$\Pi_{2}$,$\Pi_{3}$}, which is the symmetric group $S_{3}$. For example, the 4-cycle $p = (1,2,3,4)$ acts on subsets of order two as follows: $$ \{1,2\} \rightarrow \{2,3\}\\ \{1,3\} \rightarrow \{2,4\}\\ \{1,4\} \rightarrow \{1,2\}\\ \{2,3\} \rightarrow \{3,4\}\\ \{2,4\} \rightarrow \{1,3\}\\ \{3,4\} \rightarrow \{1,4\}. $$

Looking at this action, one sees that $p$ acts on the set {$\Pi_{1}$,$\Pi_{2}$,$\Pi_{3}$} of partitions as the traposition ($\Pi_{1} \Pi_{3}$) that fixes $\Pi_{2}$ and interchanges on $\Pi_{1}$ and $\Pi_{3}$.

If $p$ and $q$ are elements of $S_{4}$, the product $pq$ is the composed permutation $p . q$, and the action of the action of $pq$ on the set {$\Pi_{1}$,$\Pi_{2}$,$\Pi_{3}$} is the composition of the actions of $q$ and $p$. Therefore $\phi(pq) = \phi(p)\phi(q)$, and $\phi$ is a homomorphism.

The map is surjective, so its image is the whole group $S_{3}$. Its kernel can be computed. It is the subgroup $S_{4}$ consisting of the identity and the three products of disjoint transpositions: K = {${1,(12)(34),(13)(24),(14)(23)}$}.

The part I don't understand is how $p$ acts on subsets of order 2...

{1,2} $\rightarrow$ {2,3}

{1,3} $\rightarrow$ {2,4} etc. How do they work? Just seems random to me. Everything after this point.. I am lost on :(

Any help would be appreciated. Thanks

  • 9
    $\begingroup$ It just sends the pair $\{x,y\}$ to the pair $\{p(x),p(y) \}$. $\endgroup$ – Geoff Robinson Aug 24 '12 at 12:01

Basically, your test is just having them map the elements as an ordered set - in other words, $p:(a, \ b) \rightarrow (p(a), \ p(b))$. In the case of the examples that your book had, (1, 2, 3, 4) acts on (1, 3) in this way, because (1, 2, 3, 4) sends 1 to 2, and 3 to 4, so you get (2, 4). Hope that helps.


The map $p$ is simply how it is defined. If you're still confused on the how $p$ acts on the subsets of order $2$, i.e. the transpositions, you can think of it as a shift along two strings of $1234$. So for example $(2,4) \to (1,3)$ would be as though you did the following: $$ 1\color{red}{2}3\color{blue}{4}\color{red}{1}2\color{blue}{3}4 $$ So you just shift by $3$ positions.

  • $\begingroup$ Nice colors!!!! $\endgroup$ – Igor Rivin Dec 23 '13 at 16:45

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