Group presentations and isomorphism? I am reading Dummit and Foote, and they have only introduced group presentations very informally, so I am worried about the technicalities. 
I have to prove that the subgroup of $SL_2(\mathbb{F}_3)$ generated by 
$$A = \begin{bmatrix}
0 & -1\\
1 & 0
\end{bmatrix}
\hspace{2cm}
B = \begin{bmatrix}
1 & 1\\
1 & -1
\end{bmatrix}
$$
is isomorphic to $Q_8$. I know that one presentation of $Q_8$ is
$$Q_8 = \langle -1, i, j, k|(-1)^2=1, i^2=j^2=k^2=ijk=-1 \rangle.$$
Now, if we identify $A$ with $i$, $B$ with $j$, $AB$ with $k$, and $-1$ with $-I$ (where $I$ is the identity matrix, and I can prove that both $I$ and $-I$ are in my subgroup of matrices), then my matrices satisfy the same relations as the ones in my group presentation for $Q_8$.
My question is, am I done? Is this enough to show that the two groups are isomorphic? Or is there something I am missing? 
 A: Unfortunately, it is not quite enough.
What you have done so far is to construct a map $η\colon \{-1, i, j, k\} → \operatorname{SL}_2 \mathbb F_3$ such that $$η(i)^2 = η(j)^2 = η(k)^2 = η(i)η(j)η(k) = η(-1).$$
This is exactly how you start. By the universal property of group presentations there is now exactly one group morphism
$$Q_8 = ⟨-1, i, j, k;~i^2 = j^2 = k^2 = ijk = -1⟩ → \operatorname{SL}_2 \mathbb F_3,$$
extending $η$. So far, so good. Now you need to know that this extension is an isomorphism.
However, this would also be possible if the target was the trivial group, not $\operatorname{SL}_2 \mathbb F_3$. So why should your particular map be an isomorphism onto its image?
To show that this extension is an isomorphism onto its image, you would need to still show that it is injective.
To show it’s injective, you can either try to argue that its kernel is trivial or to argue that $Q_8$ indeed only has at most eight elements, as suggested by Derek Holt. In both cases you need to examine the elements of $Q_8$ as strings in the letters $-1, i, j, k$ and use their relations.
