Quaternion Group is Hamiltonian

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Theorem

The quaternion group $Q$ is Hamiltonian.


Proof

For clarity the Cayley table of $Q$ is presented below:

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


By definition $Q$ is Hamiltonian if and only if:

$Q$ is non-abelian

and:

every subgroup of $Q$ is normal.

$Q$ is non-abelian as demonstrated by the counter-example:

$a b \ne b a$

From Subgroups of Quaternion Group:

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.


From Trivial Subgroup and Group Itself are Normal:

$Q$ and $\set e$ are normal subgroups of $Q$.

From Center of Quaternion Group, $\gen {a^2} = \set {e, a^2}$ is the center of $Q$.

From Center of Group is Normal Subgroup, $\set {e, a^2}$ is normal in $Q$.


The remaining subgroups of $Q$ are of order $4$, and so have index $2$.

From Subgroup of Index 2 is Normal it follows that all of these order $4$ subgroups of $Q$ are normal.


That accounts for all subgroups of $Q$.

Hence the result.

$\blacksquare$


Sources

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