# An epimorphism from $S_{4}$ to $S_{3}$ having the kernel isomorphic to Klein four-group

Exercise $7$, page 51 from Hungerford's book Algebra.

Show that $N=\{(1),(12)(34), (13)(24),(14)(23)\}$ is a normal subgroup of $S_{4}$ contained in $A_{4}$ such that $S_{4}/N\cong S_{3}$ and $A_{4}/N\cong \mathbb{Z}_{3}$.

I solved the question after many calculations. I would like to know if is possible to define an epimorphism $\varphi$ from $S_{4}$ to $S_{3}$ such that $N=\ker(\varphi)$.

Thanks for your kindly help.

• How did you prove that $S_4/N\cong S_3$ without finding an epimorphism $\varphi\colon S_4\to S_3$ such that $N=\mathrm{ker}(\varphi)$? Once you have the isomorphism $f\colon S_4/N\to S_3$, let $\varphi=f\circ\pi$, where $\pi\colon S_4\to S_4/N$ is the canonical projection onto the quotient. – Arturo Magidin Feb 7 '12 at 20:12
• It's not clear what you mean by "I solved the question." If you've already shown that $S_4/N \cong S_3$, then you've shown that you can define an epimorphism from $S_4$ to $S_3$ with kernel $N$. – Thomas Andrews Feb 7 '12 at 20:14
• @ArturoMagidin: I wrote all elements of $S_{4}$, then I've found all right and left cosets of $N$ in $S_{4}$ and saw that $aN=Na$, for those permutations $a$ of $S_{4}$ that fix the element $4$. That's the way I solved the question, but it required a lot of calculations and patience. – spohreis Feb 7 '12 at 20:19
• @ThomasAndrews: Sorry, maybe I wasn't clear. I would like to exhibt such epimorphism. – spohreis Feb 7 '12 at 20:21
• @spohreis: What you describe does not sound to me like "you solved the question". Since $N$ is normal, the fact that $aN=Na$ for all $a\in S_4$ follows immediately, in particular for those elements of $a$ that fix $4$. Observing this does not, in any way, establish that there is an isomorphism between $S_4/N$ and $S_3$. – Arturo Magidin Feb 7 '12 at 20:23

## 3 Answers

Yes, there is. My favorite way of doing that is the following. There are exactly three ways of partitioning the set $\{1,2,3,4\}$ to two disjoint pairs, namely $$P_1=\{\{1,2\},\{3,4\}\},\quad P_2=\{\{1,3\},\{2,4\}\},\quad\text{and}\quad P_3=\{\{1,4\},\{2,3\}\}.$$ Now given a permutation $\sigma\in S_4$ it acts naturally on the set $\{P_1,P_2,P_3\}$ of such partitions "elementwise", and thus gives us a permutation $\overline{\sigma}\in Sym(\{P_1,P_2,P_3\})$. This correspondence $\sigma\mapsto \overline{\sigma}$ is (one of) the epimorphism(s) you are looking for.

More details: $\overline{\sigma}$ takes the partition $P_1$ to the partion $\{\{\sigma(1),\sigma(2)\},\{\sigma(3),\sigma(4)\}\}$ and similarly for the others. For example, when $\sigma=(234)$ we get that \begin{aligned} \overline{\sigma}(P_1)&=\{\{1,3\},\{4,2\}\}=P_2,\\ \overline{\sigma}(P_2)&=\{\{1,4\},\{3,2\}\}=P_3,\\ \overline{\sigma}(P_3)&=\{\{1,2\},\{3,4\}\}=P_1,\\ \end{aligned} so the resulting permutation is the 3-cycle $P_1\mapsto P_2\mapsto P_3\mapsto P_1$.

It is tedious but straightforward to check that the mapping $\sigma\mapsto \overline{\sigma}$ is surjective. It is a bit easier to check the all the permutations of the subgroup $N$ leave all the partitions $P_j,j=1,2,3,$ invariant.

• Exactly as I was taught in class of a surjective homomorphism from $S_4$ to $S_3$. – user38268 Feb 9 '12 at 21:35
• A short argument for surjectivity: For $σ = (2~3~4)$, $σP_1 = P_2 ≠ P_1$; thus $\bar σ ≠ \mathrm{id}$ with $\operatorname{ord} \bar σ \mid \operatorname{ord} σ = 3$. For $τ = (2~3)$, $τP_1 = P_2 ≠ P_1$; thus $\bar τ ≠ \mathrm{id}$ with $\operatorname{ord} \bar τ \mid \operatorname{ord} τ = 2$. Therefore, the image of the homomorphism contains elements $\bar σ$, $\bar τ$ of order $3$ and $2$ respectively, so it must be a subgroup of $S_3$ of at least order $6$, so it is all $S_3$. – k.stm May 22 '17 at 0:32

Take a tetrahedron. The symmetries form the group $S_4$ on the vertices. Consider the action of these symmetries on the three pairs of opposite edges of the tetrahedron. i.e. Pair 1 - edges 12, 34; Pair 2 edges 13, 24; pair 3 edges 14, 23.

I'll leave you to work out the details. The other platonic solids also give some geometric realisations of other relationships between groups.

Here is an approach:

Proof Idea: $S_4/N$ is a group with 6 elements. There are only two such groups, one is cyclic and the other is $S_3$, and $S_4/N$ cannot have elements of order $6$ thus must be $S_3$.

First it is easy to show that $N$ is normal in $S_4$. It follows that $S_4/N$ is a group with $6$ elements. Let us call this factor $G$.

Now, $G$ is a group of order $6$. As no element of $S_4$ has order $6$, it follows that $G$ has no element of order $6$.

Pick two elements $x,y \in G$ such that $\operatorname{ord}(x)=2$ and $\operatorname{ord}(y)=3$. Then, $e, y, y^2, x, xy, xy^2$ must be 6 distinct elements of $G$, and hence $$G= \{ e, y, y^2, x, xy, xy^2 \}$$

Now, let us look at $yx$. This cannot be $xy$, as in this situation we would have $\operatorname{ord}(xy)=6$. This cannot be $e, x, y, y^2$ either. This means that $$yx=xy^2$$

Now it is trivial to construct an isomorphism from $G$ to $S_3$.