# Quaternion Group/Subgroups

## Subgroups of the Quaternion Group

Let $Q$ denote the quaternion group, whose group presentation is given as:

$\Dic 2 = \gen {a, b: a^4 = e, b^2 = a^2, a b a = b}$

The subsets of $Q$ which form subgroups of $Q$ are:

 $\ds$  $\ds Q$ $\ds$  $\ds \set e$ $\ds$  $\ds \set {e, a^2}$ $\ds$  $\ds \set {e, a, a^2, a^3}$ $\ds$  $\ds \set {e, b, a^2, a^2 b}$ $\ds$  $\ds \set {e, a b, a^2, a^3 b}$

From Quaternion Group is Hamiltonian we have that all of these subgroups of $Q$ are normal.

## Proof

Consider the Cayley table for $Q$:

$\begin{array}{r|rrrrrrrr} & e & a & a^2 & a^3 & b & a b & a^2 b & a^3 b \\ \hline e & e & a & a^2 & a^3 & b & a b & a^2 b & a^3 b \\ a & a & a^2 & a^3 & e & a b & a^2 b & a^3 b & b \\ a^2 & a^2 & a^3 & e & a & a^2 b & a^3 b & b & a b \\ a^3 & a^3 & e & a & a^2 & a^3 b & b & a b & a^2 b \\ b & b & a^3 b & a^2 b & a b & a^2 & a & e & a^3 \\ a b & a b & b & a^3 b & a^2 b & a^3 & a^2 & a & e \\ a^2 b & a^2 b & a b & b & a^3 b & e & a^3 & a^2 & a \\ a^3 b & a^3 b & a^2 b & a b & b & a & e & a^3 & a^2 \end{array}$

We have that:

$a^4 = e$

and so $\gen a = \set {e, a, a^2, a^3}$ forms a subgroup of $Q$ which is cyclic.

We have that:

$b^2 = a^2$

and so $\gen b = \set {e, b, a^2, a^2 b}$ forms a subgroup of $Q$ which is cyclic.

We have that:

$\paren {a b}^2 = a^2$

and so $\gen {a b} = \set {e, a b, a^2, a^3 b}$ forms a subgroup of $Q$ which is cyclic.

We have that:

$\paren {a^2}^2 = e$

and so $\gen {a^2} = \set {e, a^2}$ forms a subgroup of $Q$ which is also a subgroup of $\gen a$, $\gen b$ and $\gen {a b}$.

That exhausts all elements of $Q$.

Any subgroup generated by any $2$ elements of $Q$ which are not both in the same subgroup as described above will generate the whole of $Q$.

$\blacksquare$