$S_4/H \simeq S_3$ where $H$ is a normal subgroup

Prove that the group of permutations of four symbols $S_4$ contains a normal subgroup H such that the quotient group $S_4/H$ is isomorphic to the group of permutations of three symbols $S_3$.

$S_4$ has order $24$. Any subgroup thus will have order $1, 2, 3, 4, 6, 8, 12$ or $24$.

A normal subgroup is a union of conjugacy classes which in this case correspond to the cycle shapes as follows:

1. 6 of the form $(abcd)$
2. 8 of the form $(abc)(d)$
3. 3 of the form $(ab)(cd)$
4. 6 of the form $(ab)(c)(d)$
5. 1 identity element

If $S_4/H\simeq S_3$ with one of the above normal subgroups as $H$ then by Lagrange $|H|=\frac{|S_4|}{|S_3|}=4$ but there are no normal subgroups with this order.

Would you be able to help me with this?

• The union of conjugacy classes $3$ and $5$ forms a subgroup. – Ravi May 29 '16 at 19:52
• Why do you think there are no normal subgroups with this order? Think of groups of order 4, any group of order 4 is either isomorphic to $V_4$ or $C_4$, find subgroups of $S_4$ that are isomorphic to one of these two and check whether they are normal – M. Van May 29 '16 at 19:53
• Do also have a look at this answer – Johannes Huisman May 29 '16 at 20:26

1) Check that a subgroup $\;H\;$ of a group is a normal subgroup iff it is the (disjoint) union of conjugacy classes.

2) With (1), or directly, check that $\;V:=\left\{\,(1),\,(12)(34),\,(13)(24)\,,\,(14)(23)\,\right\}\lhd S_4\;$

3) Take now the quotient group $\;S_4/V\;$ . By Lagrange's theorem, this group's order is six, so it is either $\;S_3\;$ or the cyclic group $\;C_6\;$ . In order to show it is actually the former it is enough to show there are two different elements of order two in the quotient (be sure you can justify this), and indeed:

$$(12)V\neq V\;,\;\;\text{and}\;\;\left((12) V\right)^2=\overline1=V\,,\,\,\text{and the same's true for}\;\;(13)V$$

Now, the above two elements are the same iff:

$$(12)V=(13)V\iff (12)^{-1}(13)\in V\iff (12)(13)=(132)\in V$$

and since $\;(132)\notin V\;$ we've finished.

A normal subgroup, as you noted, is a union of conjugacy classes. It must also contain the identity element, which accounts for $1$ element of $H$. Do the sizes of any of the other conjugacy classes plus $1$ give you $4$?

Edit: Now that you know the correct subgroup, here's something to think about for proving $S_4/H\simeq S_3$. Note that every element of $S_4$ can be written as a product of transpositions. There are $6$ transpositions: $(12),(13),(14),(23),(24),(34)$. In the quotient, we have, for instance, $(12)(34)H=H$, so $(12)H=(34)H$. Similarly, $(13)H=(24)H$, and $(14)H=(23)H$. So in the quotient $S_4/H$ everything can be written as a product of $(12)H$, $(13)H$, and $(23)H$.

Now using that fact, can you define a surjective homomorphism $S_4\to S_3$ with kernel $H$? (Note that everything in $S_3$ is a product of $(12),(13),(23)$. Since $(12)H=(34)H$ in the quotient, your map should send $(12)$ and $(34)$ to the same place. Similarly with the other cosets that are equal.)

• Oh ok, so could we let $H=\{e, (ab)(cd)\}$? This would have the required order $4$ but how can we prove that $S_4/H \simeq S_3$? – amiz9 May 29 '16 at 19:57
• See the added details. – kccu May 29 '16 at 20:10
• @amiz9 Why do you think your $\;H\;$ has order four? It has only two elements as far as I can see... – DonAntonio May 29 '16 at 21:05
• @Joanpemo I think I am confusing it all up – amiz9 May 29 '16 at 21:08
• @Joanpemo I took $(ab)(cd)$ to mean all elements of that form, i.e., $(12)(34)$, $(13)(24)$, and $(14)(23)$. – kccu May 29 '16 at 21:34