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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 $G$?

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    $\begingroup$ Just write out all elements you can form with $S$. At some point they are no new ones. Here with $Q_8$ this happens quite quickly. Then google "generating sets of quaternion group", to see if you were right (see the answer in this duplicate). $\endgroup$ – Dietrich Burde Oct 16 '16 at 20:36
  • $\begingroup$ @DietrichBurde ok, i will try it. $\endgroup$ – False Promise Oct 16 '16 at 20:42
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You can construct the multiplication table of the group. You can do that graphically, too, by constructing the Cayley graph. Like this one:

Cayley graph of the quaternion group

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.

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  • $\begingroup$ This is a nice picture (+1). OK, it is Wikipedia, but still... $\endgroup$ – Dietrich Burde Oct 16 '16 at 20:45
  • $\begingroup$ When i should stop, when i'm forming a multiplication table of a group? $\endgroup$ – False Promise Oct 16 '16 at 20:46
  • $\begingroup$ When the table is complete. Try constructing it, inserting one element at a time. If you use only 1, then you obtain a trivial $1\times1$ table (the trivial subgroup). When you add $i$, you obtain a $4\times4$ table (the group $\{1,-1,i,-i\}$ Then you can add $j$... $\endgroup$ – zar Oct 16 '16 at 20:52
  • $\begingroup$ @DietrichBurde, should I make a link to Wikipedia? Or a note? $\endgroup$ – zar Oct 16 '16 at 20:53
  • $\begingroup$ @zar So in general case, if i have 3 elements, then my table is $3\times 3$, if 4, then it's $4\times 4$ and so on? $\endgroup$ – False Promise Oct 16 '16 at 20:56

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