Quaternion group Let $S = \{ i, j, k\}$, $G =  \langle S  \rangle$. And  $i\cdot j = k $,    $j\cdot k = i, k\cdot i = j, i^2=j^2=k^2$.
I have to show that $G = \{e, i, j, k, i^2, i^2\cdot i, i^2\cdot j, i^2\cdot k \}$.
So my plan is to show that every arbitrary element from $G$ can be expressed in one of the following forms:$(i^2)^m, (i^2)^m\cdot i, (i^2)^m\cdot j,(i^2)^m\cdot k$, where $m$ is non-negative integer. And then i can prove $(i^2)^2 = e$. Unfortunately, i don't know what to start with, because there are a lot of representations of my arbitraty element.
And in general case, if i know some generating set $\langle S  \rangle$, how can i find $G$?
 A: You can construct the multiplication table of the group. You can do that graphically, too, by constructing the Cayley graph. Like this one:

The red arrow corresponds to the multiplication by $i$, the green arrow to the multiplication by $j$.
[Edit: expanded table creation]
Start with the set $\{1\}$. Obviuosly, if you try to create the multiplication table, you obtain something like this:
$$
\begin{array}{|r|r|}
\hline
\cdot & 1 \\ \hline
1 & 1 \\ \hline
\end{array}
$$
Now you would like to add $-1$. This is what you get:
$$
\begin{array}{|r|r|r|}
\hline
\cdot & 1 & -1\\ \hline
1 & 1 & -1\\ \hline
-1 & -1 & 1\\ \hline
\end{array}
$$
Now, if you try to add $i$:
$$
\begin{array}{|r|r|r|r|}
\hline
\cdot & 1 & -1 & i\\ \hline
1 & 1 & -1 & i\\ \hline
-1 & -1 & 1 & -i\\ \hline
i & i & -i & -1\\ \hline
\end{array}
$$
you realize that this table is not complete, as it contains a new element: $-i$. So you have to add also $-i$ to your set:
$$
\begin{array}{|r|r|r|r|r|}
\hline
\cdot & 1 & -1 & i & -i\\ \hline
1 & 1 & -1 & i & -i\\ \hline
-1 & -1 & 1 & -i & i\\ \hline
i & i & -i & -1 & 1\\ \hline
-i & -i & i & 1 & -1\\ \hline
\end{array}
$$
and now the table is complete.
Now it's time to add $j$, but as you start creating the table, you realize that you have to add also $-j$ (because $-1\cdot j=-j$), $k$ (because it's equal to $i\cdot j$) and $-k$. Now the set $\{1,-1,i,-i,j,-j,k,-k\}$ is closed.
