Classifying the Groups of order 8 (Exact Sequence)

Question

I need help understanding this proof of the classification of groups of order 8. I understand the case where all the non-trivial elements have order 2. The next case is when the group has an element of order 4.

Let $G$ be a group of order 8, and let $g \in G$ be an element of order $4$. Let $H = \left< g \right>$. $H$ has index 2 in $G$, so $H$ must be normal. From here, we obtain an exact sequence $$1 \rightarrow \mathbb{Z}/4\mathbb{Z} \rightarrow G \rightarrow \mathbb{Z}/ 2\mathbb{Z} \rightarrow 1.$$

Pick $h \in G$ whose image is non-trivial in $\mathbb{Z}/2\mathbb{Z}$. We know $h^2 \in \mathbb{Z}/4\mathbb{Z}$ so $h^2 = 1, g, g^2, g^3$. We know that $hgh^{-1}$ generates $G$, so we must have $hgh^{-1} = a, a^3$.

Then, we run through the possible values of $h^2, hgh^{-1}$ and check if they form a group.

• $h^2 = 1, hgh^{-1} = g$ gives $\mathbb{Z}/2\mathbb{Z} \times \mathbb{Z}/4\mathbb{Z}$.
• $h^2 = g, hgh^{-1} = g$ gives $\mathbb{Z}/8\mathbb{Z}$.
• $h^2 = g^2, hgh^{-1} = g$ gives $\mathbb{Z}/8\mathbb{Z}$.
• $h^2 = g^3, hgh^{-1} = g$ gives $D_8$.
• $h^2 = 1, hgh^{-1} = g^3$ gives $D_8$.
• $h^2 = g, hgh^{-1} = g^3$, the group collapses.
• $h^2 = g^2, hgh^{-1} = g^3$ gives $Q_8$.
• $h^2 = g^3, hgh^{-1} = g^3$ gives $\mathbb{Z}/2\mathbb{Z} \times \mathbb{Z}/4\mathbb{Z}$.

My issues:

1. I don't really get where the exact sequence comes from. Don't we need to specify homomorphisms between the groups?
2. What does it mean for the image of $h$ to be non-trivial? The image under what?
3. Why is $h^2 \in \mathbb{Z}/4\mathbb{Z}$? The answer to this probably stems from the previous one.
4. How do we know $hgh^{-1}$ generates $G$?

Answer 1

First there is an error - you should have $hgh^{-1}$ generates $H$.

1. Whenever you have a normal subgroup $H$ of $G$ you have an exact sequence $$1\to H\to G\to G/H\to 1$$ where $H$ embeds into $G$ via $g\mapsto g$ and $G$ projects onto $G/H$ via $g\mapsto gH$.

2. Following above the image of $h$ is nontrivial if $h\notin H$.

3. As $|G|/|H|=2$, $G/H\cong\mathbb{Z}/2\mathbb{Z}$. Therefore $H=(hH)^2=h^2H$ which means $h^2\in H$.

4. As $g$ generates $H$, we have $H=\{1,g,g^2,g^3\}$. But $H$ is normal in $G$ so $$H=hHh^{-1}=\{1,hgh^{-1},hg^2h^{-1},hg^3h^{-1}\}$$ so $hgh^{-1}$ generates $H$.