TITLE: Classifying the Groups of order 8 (Exact Sequence)
QUESTION [1 upvotes]: 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: 

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

REPLY [1 votes]: First there is an error - you should have $hgh^{-1}$ generates $H$.

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 $g\mapsto gH$.
Following above the image of $h$ is nontrivial if $h\notin H$.
As $|G|/|H|=2$, $G/H\cong\mathbb{Z}/2\mathbb{Z}$. Therefore $H=(hH)^2=h^2H$ which means $h^2\in H$.
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 $H$.
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