Show that $\{1, (1\quad3)(2\quad4), (1\quad4)(2\quad3),(1\quad2)(3\quad4)\}$ is a normal subgroup of $\mathrm{S}_4$

Question

Show that $\{1, (1\quad3)(2\quad4), (1\quad4)(2\quad3),(1\quad2)(3\quad4)\}$ is a normal subgroup of $\mathrm{S}_4$

I could do this by brute force, but I'm wondering if there are any observations that would let me avoid that approach?

Answer 1

Conjugation by any element of $S_4$ preserves the cycle structure of the element being conjugated. For example, $\sigma(13)(24)\sigma^{-1}$ is another permutation of the form $(ab)(cd)$, no matter what $\sigma \in S_4$ we choose.

Since your subset contains all of the permutations of the form $(ab)(cd)$ and nothing else except the identity, it is closed under conjugation. Since your subset is also a subgroup, this means that it is a normal subgroup.

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Edit:

Here is a more sophisticated argument which uses the fact that $S_4$ contains isomorphic copies of $D_8$ (the dihedral group of a square), along with a bit of Sylow theory.

We can view the elements $1,2,3,4$ as vertices of a square in three different ways, depending on the order in which we number the vertices.

• Square 1 has vertices numbered 1,2,3,4 in that order
• Square 2 has vertices numbered 1,2,4,3 in that order
• Square 3 has vertices numbered 1,3,2,4 in that order

The other three possible numbering sequences are simply the reverse of one of the above, which can be achieved geometrically by flipping one of the above squares upside-down.

The dihedral group of each of these squares is a subgroup of $S_4$ of order $8$. As $|S_4| = 24 = 8\cdot 3$, these dihedral groups are $2$-sylow subgroups of $S_4$. Moreover, these are all of the $2$-sylow subgroups, because by Sylow counting, the number of $2$-sylow subgroups must be a divisor of $3$.

Now, observe that $H = \{1, (13)(24), (14)(23), (12)(34)\}$ is contained in all three of the dihedral groups (to see this, it helps to use three square pieces of paper with the vertices labeled as described above; in each case, two of the operations correspond to flips about the horizontal and vertical axes, and the third operation is a rotation by 180 degrees).

Therefore $H$ is contained in the intersection of the three dihedral groups. Since these are three distinct groups of order $8$ and $H$ has order $4$, it follows that $H$ must in fact be equal to the intersection of the three dihedral groups. Since the dihedral groups are all conjugate (they are Sylow subgroups of the same order), this means that $H$ is equal to $\bigcap_{g \in G}gDg^{-1}$ where $D$ is any one of the dihedral subgroups. In other words, $H$ is the core of each $D$. Since the core of any subgroup is always a normal subgroup of the full group, the result follows.