# What are the subgroups of S_4 and D_4 ?

Aug 27, 2016

See explanation...

#### Explanation:

Note that I assume below that by ${D}_{4}$ you are using the geometric convention of the group of symmetries of a square, i.e. the dihedral group of order $8$. The same name is used differently in abstract algebra to refer to the dihedral group of order $4$ (i.e. the symmetry group of the digon), isomorphic to ${C}_{2} \times {C}_{2}$.

What I find interesting about this question is that while ${S}_{4}$ can be thought of as the symmetry group of the regular tetrahedron and ${D}_{4}$ as that of a square, ${D}_{4}$ is actually a subgroup of ${S}_{4}$.

${S}_{4}$ consists of all permutations of $4$ objects.

If we consider the vertices of a regular tetrahedron then any of its symmetries permutes its vertices. What is slightly less obvious is that any permutation of the four vertices is possible - if you allow both rotations and reflection.

If we reflect in a plane containing one edge and the midpoint of the opposite edge, then it transposes two vertices and leaves the other two fixed. There are $6$ such planes that we could choose, allowing us to transpose any pair of vertices. This gives us $6$ subgroups isomorphic to ${C}_{2}$ using reflection (there are others).

Since the transposition of one pair of vertices is independent of the transposition of the other two, we have $3$ subgroups isomorphic to ${C}_{2} \times {C}_{2}$.

If we only allow rotations, then we get the group of even permutations of the $4$ vertices, which is ${A}_{4}$, the alternating group of four elements.

Speaking of rotations, note that if we rotate about an axis through one vertex and the centre of the opposite face then one of the vertices is fixed and the other $3$ cyclically permuted, hence a subgroup ${C}_{3}$. Combined with reflection in a plane containing the fixed vertex, this gives us a subgroup ${D}_{3} = {S}_{3}$ - the full group of symettries of an equilateral triangle.

If we rotate about an axis through the centre of one edge and the centre of the opposite edge, then it transposes two pairs of vertices, hence a subgroup ${C}_{2}$.

If we label the $4$ vertices $a , b , c$ and $d$, then we can pick out a symmetry of the tetrahedron that cyclically permutes the four vertices. $\left(a , b , c , d\right) \mapsto \left(b , c , d , a\right)$. On its own this generates a subgroup isomorphic to ${C}_{4}$. We can pick out another operation that transposes $a$ and $c$: $\left(a , b , c , d\right) \mapsto \left(c , b , a , d\right)$. These two operations combined generate a subgroup isomorphic to ${D}_{4}$. Looking back, this means that ${D}_{4}$ has two non-trivial cyclic subgroups ${C}_{4}$ and ${C}_{2}$.

So we have identified subgroups of ${S}_{4}$:

• ${S}_{4}$
• ${A}_{4}$
• ${D}_{4}$
• ${D}_{3} = {S}_{3}$
• ${C}_{2} \times {C}_{2}$
• ${C}_{4}$
• ${C}_{3}$
• ${C}_{2} = {S}_{2}$
• ${C}_{1} = {S}_{1}$

The subgroups of ${D}_{4}$ are:

• ${D}_{4}$
• ${C}_{4}$
• ${C}_{2} = {S}_{2}$
• ${C}_{1} = {S}_{1}$