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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.

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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.2012-02-07
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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 $N$.2012-02-07
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    @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.2012-02-07
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    @ThomasAndrews: Sorry, maybe I wasn't clear. I would like to exhibt such epimorphism.2012-02-07
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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 $S_3$.2012-02-07
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    @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?2012-02-12
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    @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 $N$).2012-02-12
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    @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 $aN=Na$.2012-02-12
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    @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.2012-02-12

3 Answers 3

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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.

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    Exactly as I was taught in class of a surjective homomorphism from $S_4$ to $S_3$.2012-02-09
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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 $S_3$.2017-05-22
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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.

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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 $S_3$.