Dihedral Group D4/Subgroups
Subgroups of the Dihedral Group $D_4$
Let the dihedral group $D_4$ be represented by its group presentation:
- $D_4 = \gen {a, b: a^4 = b^2 = e, a b = b a^{-1} }$
The subsets of $D_4$ which form subgroups of $D_4$ are:
| \(\ds \) | \(\) | \(\ds D_4\) | ||||||||||||
| \(\ds \) | \(\) | \(\ds \set e\) | ||||||||||||
| \(\ds \) | \(\) | \(\ds \set {e, a, a^2, a^3}\) | ||||||||||||
| \(\ds \) | \(\) | \(\ds \set {e, a^2}\) | ||||||||||||
| \(\ds \) | \(\) | \(\ds \set {e, b}\) | ||||||||||||
| \(\ds \) | \(\) | \(\ds \set {e, b a}\) | ||||||||||||
| \(\ds \) | \(\) | \(\ds \set {e, b a^2}\) | ||||||||||||
| \(\ds \) | \(\) | \(\ds \set {e, b a^3}\) | ||||||||||||
| \(\ds \) | \(\) | \(\ds \set {e, a^2, b, b a^2}\) | ||||||||||||
| \(\ds \) | \(\) | \(\ds \set {e, a^2, b a, b a^3}\) |
Proof
Consider the Cayley table for $D_4$:
- $\begin{array}{l|cccccccc} & e & a & a^2 & a^3 & b & b a & b a^2 & b a^3 \\ \hline e & e & a & a^2 & a^3 & b & b a & b a^2 & b a^3 \\ a & a & a^2 & a^3 & e & b a^3 & b & b a & b a^2 \\ a^2 & a^2 & a^3 & e & a & b a^2 & b a^3 & b & b a \\ a^3 & a^3 & e & a & a^2 & b a & b a^2 & b a^3 & b \\ b & b & b a & b a^2 & b a^3 & e & a & a^2 & a^3 \\ b a & b a & b a^2 & b a^3 & b & a^3 & e & a & a^2 \\ b a^2 & b a^2 & b a^3 & b & b a & a^2 & a^3 & e & a \\ b a^3 & b a^3 & b & b a & b a^2 & a & a^2 & a^3 & e \end{array}$
We have that:
- $a^4 = e$
and so $\gen a = \set {e, a, a^2, a^3}$ forms a subgroup of $D_4$ which is cyclic.
We have that:
- $\paren {a^2}^2 = e$
and so $\gen {a^2} = \set {e, a^2}$ forms a subgroup of $D_4$ which is cyclic, and also a subgroup of $\gen a$.
We have that:
- $b^2 = e$
and so $\gen b = \set {e, b}$ forms a subgroup of $D_4$ which is cyclic.
We have that:
- $\paren {b a}^2 = e$
and so $\gen {b a} = \set {e, b a}$ forms a subgroup of $D_4$ which is cyclic.
We have that:
- $\paren {b a^2}^2 = e$
and so $\gen {b a^2} = \set {e, b a^2}$ forms a subgroup of $D_4$ which is cyclic.
We have that:
- $\paren {b a^3}^2 = e$
and so $\gen {b a^3} = \set {e, b a^3}$ forms a subgroup of $D_4$ which is cyclic.
Then we have that:
- $b a^2 = a^2 b$
and so $\gen {a^2, b} = \set {e, a^2, b, b a^2}$ forms a subgroup of $D_4$ which is not cyclic, but which has subgroups $\set {e, a^2}$, $\set {e, b}$, $\set {e, b a^2}$.
Then we have that:
- $b a^3 = a^2 b a$
and so $\gen {a^2, b a} = \set {e, a^2, b a, b a^3}$ forms a subgroup of $D_4$ which is not cyclic, but which has subgroups $\set {e, a^2}$, $\set {e, b}$, $\set {e, b a^2}$.
That exhausts all elements of $D_4$.
Any subgroup generated by any $2$ elements of $Q$ which are not both in the same subgroup as described above generate the whole of $D^4$.
$\blacksquare$