Prove that the group $Z_2$ x $Z_4$ is not isomorphic to $Q_8$, the quaternion group of order 8.

Question

I have never heard of the quaternion group until today by google searching. It has the elements {1, i, j, k, -i, -j, -k, -1} and some special operations. I have a couple of these exercises using the quaternion group. A little help would be great.

Answer 1

Note that the multiplication table of $Q_8$ is $$ \begin{array}{r|rrrr} & 1 & i & j & k \\ \hline 1 & 1 & i & j & k \\ i & i & -1 & k & -j \\ j & j & -k & -1 & i \\ k & k & j & -i & -1 \end{array} $$ This table immediately implies that $Q_8$ is not abelian and hence not isomorphic to $\Bbb Z_2\times \Bbb Z_4$.

Of course, we are utilizing the following lemma:

Lemma. Let $\phi:G\to H$ be a group isomorphism where $G$ is abelian. Then $H$ is abelian.

Proof. Let $h_1,h_2\in H$. Then there are $g_1,g_2\in G$ satisfying $\phi(g_i)=h_i$. It follows that $$ h_1h_2=\phi(g_1)\phi(g_2)=\phi(g_1g_2)=\phi(g_2g_1)=\phi(g_2)\phi(g_1)=h_2h_1 $$ Hence $H$ is abelian. $\Box$

Answer 2

Another argument is that there are 2 elements of order $2$ in $\mathbb{Z}_2 \times \mathbb{Z}_4$: $(0, 2)$ and $(1, 2)$, but there is only 1 element of order $2$ in $\mathbb{Q}_8$: $-1$.