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

Question

This is the proof from class:

$Q_8 = \langle a,b \mid a^4, a^2=b^2, b^{-1}ab=a^{-1} \rangle$ has a unique element of order 2, so every subgroup of $Q_8$ of order 4 must be cyclic and has $a^2$ as its element of order 2. So $N\cap H \neq 1$.

My question is, why can we conclude that every subgroup of $Q_8$ of order 4 must be cyclic and has $a^2$ as its element of order 2?

Answer 1

The only groups of order $4$ are cyclic, and Klein four-groups. A Klein four-group has three elements of order $2$, so there aren't any inside $Q_8$.

A cyclic group of order $4$ has one element of order $2$ (the square of its generator) so one inside $Q_8$ must contain $a^2$ as an element.

Answer 2

If a subgroup of $Q_8$ has order four and is not cyclic, then it is isomorphic to $\mathbb{Z}/2\mathbb{Z} \times \mathbb{Z}/2\mathbb{Z}$, and therefore would contain more than one element of order two, a contradiction.