In this document, there is a classification of all groups of order 8:
http://www2.lawrence.edu/fast/corrys/Math300/8Groups.pdf
I understood it all until the part in the third page that says:
"$b^2$ must have order $2$, so that $b \in H = \langle a \rangle $"
Why must $b$ be in the group generated by $a$? This isn't obvious to me.
Can someone help me? Thanks.
The relevant paragraph (named "Subcase 2b") has the assumption that every element of $G-H$, and therefore specifically $b$, has order $4$. Since $b$ has order $4$, $b^2$ has order $2$. Since the order of $b^2$ is not $4$, we must have $b^2 \notin (G-H)$ by the standing assumption. This means $b \in H = \langle a \rangle$.