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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.

Mike Pierce
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    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 07 '12 at 20:12
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    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 07 '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. –  Feb 07 '12 at 20:19
  • @ThomasAndrews: Sorry, maybe I wasn't clear. I would like to exhibt such epimorphism. –  Feb 07 '12 at 20:21
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    @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 07 '12 at 20:23
  • @ArturoMagidin: I was thinking about something else. Let $a_{i},,,i\in{1,2,3,4,5,6}$ be those elements of $S_{4}$ that fix $4$. I proved that $S_{4}=\cup_{i=1}^{6}a_{i}N$ and $a_{i}N=Na_{i}$. If $g\in S_{4}$ then $a\in a_{1}N$ for some $i\in {1,2,3,4,5,6}$. Thus $aN=a_{i}N=Na_{i}=Na$ since $a\in Na_{i}$. What do you think about that know? –  Feb 12 '12 at 12:44
  • @spohreis: I still think I don't understand what it is you think you are proving. If $G$ is any group, and $N$ is any normal subgroup, then for every $g$ we have $gN=Ng$. So why are you "proving" that here? The rest just seems to be "every element of $S_4$ is equivalent to some element that fixes $4$ modulo $N$." Fine, that's necessary for the isomorphism, but it's not sufficient as stated (you would need to show that the set of elements that fix $4$ is a subgroup, and that they form a list of coset representatives, i.e., no two are equivalent modulo $N$). – Arturo Magidin Feb 12 '12 at 20:44
  • @ArturoMagidin: Because I don't know if $N$ is a normal subgroup of $S_{4}$. I am proving that those elements of $S_{4}$ that fix the element $4$ form a complete list of coset representatives and for any $a\in S_{4}$ we have $aN=Na$. –  Feb 12 '12 at 21:13
  • @spohreis: Why don't you know that $N$ is normal? It consists of a union of conjugacy classes (all trivial permutations, all products of two transpositions). And still, what you did does not establish what you want, since you did not argue why it was a complete set of coset representatitives, nor did you show that the set of permutations fixing $4$ is a subgroup. At best, it is a rather incomplete effort, and it is also rather jumbled. – Arturo Magidin Feb 12 '12 at 21:36

3 Answers3

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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.

Jyrki Lahtonen
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    Exactly as I was taught in class of a surjective homomorphism from $S_4$ to $S_3$. –  Feb 09 '12 at 21:35
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    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 00:32
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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.

Mark Bennet
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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 $S_3$.

Jyrki Lahtonen
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N. S.
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