Does there exist some isomorphism from $Q_8$ to $\Bbb Z_4 \times\Bbb Z_2$?

Question

I am wondering if some direct decomposition exists for quaternion group. I think that I am mixing some things, but let me explain and ask for clarification, tips from your side to let me understand my misconception.

First of all we have first isomorphism theorem which says that: $ H \cong G / \ker(f) $ so can I threat it as $ G \cong H \times \ker(f) $ if the $ \ker(f) $ create a group itself?

In $Q_8$ we could choose $ H $ as $ \{1, -1, i, -i\} $ which results that I have to take $ \ker(f) $ such as $ \{j, k \} $ which do not form a group itself or maybe I should choose $ \mathbb Z_2 $ which is a group so then I can claim that $ Q_8 \cong \{1, -1, i, -i\} \times \mathbb Z_2 $?

Answer 1

Since $\Bbb Z_4\times\Bbb Z_2$ is abelian whereas $Q_8$ isn't, they are not isomorphic. But, yes, $\{1,-1,i,-i\}$ is a normal subgroup of $Q_8$ which is isomorphic to $\Bbb Z_4$, and $Q/\{1,-1,i,-i\}\simeq\Bbb Z_2$.

Answer 2

The quaternion group is not abelian, because for example $ij=-ji$. So it cannot be isomorphic to an abelian group. On the other hand, there is indeed a short exact sequence $$ 1 \rightarrow \{\pm 1,\pm i\}\rightarrow Q_8\rightarrow C_2\rightarrow 1, $$ with $Q_8/\{\pm 1,\pm i\}\cong C_2$, but it is not split, i.e., $Q_8$ is not a semidirect product of $C_4$ and $C_2$ - and in particular not a direct product.

Reference: $Q_8$ is NOT the semidirect product of a group of order 4 and a group of order 2