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 $D^4$.

$\blacksquare$