Presentation of a group question So I know that given a presentation of a group $G$, one can derive from the relations of the group presentation any element in the group $G$ right. However, I do have some confusion.
If we take $G=Q_8$ , the famous non-commutative group (Quaternion group), we know that the group presentation of $Q_8$ is as follows 
$$\large{Q_8 = \big<i,j \space  \big| \space i^4 = 1, j^2 = i^2, j^{-1}ij = i^{-1} \big>}$$ 
Now it is argued that one can find any element of the group $Q_8$ using the relations given in a group presentation.
Now I want to find $k$ because I do know that $k \in Q_8$, and I know that $ij = k$. How can I find $ij$ from these relations.
Here is my attempt 
$i^4=1 \implies i^2i^2 = 1$and since we know that $i^2 =j^2$ so we get that $i^2i^2= 1 \implies i^2j^2 =1$. Now I left multiply by $i^{-1}$ and right multiply by $j^{-1}$ both sides of the equation to end up with $ij = i^{-1}j^{-1}$ and since I know that $i^{-1} = j^{-1}ij$ then I get $ij = j^{-1}ijj^{-1}$ and hence I get $ij = j^{-1}i$.
The thing is I never find $k$. Basically my question is the following

Given a group presentation ,how come can we find all the elements using the relations in a group presentation even though all the elements may not be listed in the the group presentation. You see there is no $k$ what so ever in the group presentation of $Q_8$. So how come I will be able to find it ?  So should we make life easier, and include all the elements in a group presentation. Even if we did that, there is no clue that $ij = k$ in that group presentation, One must know before hand this relation or else how can he even derive it. Well he can get $ij$ but then he may not be able to know what is $ij$ equals anyway`

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I got this from dummit book page(219) in the section A word on free groups (section 6.3)
Thanks for taking the time to read my question :)
 A: Well, the thing is, you did find $k$ -- it just didn't pop out named $k$, so to speak. Everything you know about $k$ will be just as true for $ij$ here, it just looks a little different (but only superficially).
From the presentation, you know that $ij$ is in the group being presented, because it is a group, and must be closed under the group operation.
As Matt Samuel said in the comments, presentations aren't always nice - and often they're pretty demanding, computationally. But, with sheer willpower, you can often wrestle such a presentation to learn more about your group. Here's the way I would approach it, but I make no claims of efficiency!
Disclaimer: Working with presentations is like an exercise in using algebraic identities dozens of times; I find such a thing impossibly hard to follow looking at someone else's work, but moderately fun to do once or twice a year. So, you'll probably want to follow along with a pencil and paper, if you actually want to digest any of this.
Since $i^4 = 1$, we must have that $\langle i \rangle = \{1, i, i^2, i^3\}$ is a cyclic subgroup. And, since $i^2 = j^2$, we can verify that $\langle j \rangle$ is another cyclic subgroup of size four. Except, in addition, we can clean up by using a few relations, to see that $\langle j \rangle = \{1, j, j^2, j^3\} = \{1, j, i^2, i^2j\}$.
So far, we've identified six unique elements: $1, i, j, i^2, i^3,$ and $i^2j$. But, we haven't learned much about $ij$, so we'll at least find its order now. Note that by the relation $j^{-1}ij = i^{-1}$, we know that $iji = j$ (right multiply by $i$, and left multiply by $j$). Now
$$(ij)^2 = \underbrace{iji}_j\cdot j = j^2 = i^2,$$
and by the same reasoning above, we know that $\langle ij \rangle = \{1, (ij), (ij)^2, (ij)^3\} = \{1, ij, i^2, i^3j\}$.
So now our list of elements contains
\begin{array}{cccc} 1 & i & i^2 & i^3 \\ j & ij & i^2j & i^3j \end{array} at least. Now, in order to show that we've found everything, we'll have to show that multiplication (on either side) by $i$ or $j$ gives us back something from this list. Of course, left multiplication by $i$, and right multiplication by $j$, will easily be seen to give us back something from the list.
In general, it would be nice to be able to put all of our products in the form $i^mj^n$, at which point we'll show that $m \in \{0, 1, 2, 3\}$ and $n \in \{0, 1\}$, and our list is indeed exhaustive. In order to arrive at this form, the commutator $[j, i] = j^{-1}i^{-1}ji$ will be extremely helpful, as $ij[j, i] = ji$. We'll see that $[j,i]$ will allow us to let the $i$'s and $j$'s switch places if we're willing to pay the commutator price. We'll see that here, $[j, i] = i^2 = j^2 = (ij)^2$. Try it yourself, what follows is one way.
Since $j^{-1}i^{-1} = (ij)^{-1} = i^3j$, we have
$$j^{-1}i^{-1}ji = i^3jji = i^3j^2i = i^3i^2i = i^2.$$
In practice, this is quite useful. Let's pick $i^2j$ from our list above, and right multiply by $i$, attempting to simplify $i^2ji$ to give it the form $i^mj^n$. Since we know that $ji = ij[j,i] = iji^2 = ijj^2 = ij^3$, we have
$$i^2ji = i^2(ij^3) = i^3j^3 = i^{-1}j^2j = i^{-1}i^2j = ij,$$ something from our list.
I'll spare you the rest (verifying the list is truly exhaustive), because it's really more enlightening just to dig in and see what you can figure out. But hopefully it will give you more of an idea how such a thing can work, in practice. You could take it a step further and construct the multiplication table, or verify all the things you know about $Q_8$, using the element list here.
For some groups, presentations are really nice; you can basically collapse the multiplication table of the dihedral groups into a $2 \times 2$ table (reflections and rotations), using presentations.
