You can certainly achieve a rotation of side as a composition of moves of the other sides only. The most instructive way to see this is to take your favorite algorithm for solving the cube from an arbitrary position and figure out how to modify it such that it never turns the first side you solve. (Whether this is easy depends on what your favorite algorithm is, of course. My own usual one is almost there except for the initial-cross stage and the final turning of the top corners, both of which are easily replaced). Then turn your chosen side by one quarter, and use the modified algorithm to solve from that position. Whatever moves you make during the solve will add up to a "turn this side without turning it" combination.
Also, 5 is the minimum number of single-side turns that generate the group, because restricting movement to any set of 4 sides will either leave one edge unmovable or be unable to flip an edge to the opposite orientation.
A full set of relations between 5 basic quarter-turns would be absolute horrible, though.
If the generators are not required to be simple moves, we don't need as many as 5 of them. It's easy to get down to 3 quite complex operations:
- $\alpha$, a cyclic permutation of 11 of the 12 edges simultaneously with 7 of the 8 corners.
- $\beta$, swaps the edge that $\alpha$ doesn't touch with another one, and swaps the corner that $\alpha$ doesn't touch with another one.
- $\gamma$, turns two corners and flips two edges.
Here there's some hope of being able to write down some reasonably natural relations, but I haven't tried to see whether the details work out.
I'm not sure if we can get down to two generators -- could the role of $\gamma$ be folded into $\beta$, perhaps? Edit: The next time the question came up (and I had forgotten this one existed) I tried anew and came up with a blueprint for a two-generator solution: Minimal generating set of Rubik's Cube group
One generator is of course not possible, since the group is not abelian.
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.
Best Answer
$\mathbb Z$ is a free group. We can let $S=\{1\}$ and then indeed have the defining property of free group: For any group $G$ and map $f\colon S\to G$ there exists one and only one group homomoprhism $\phi\colon\mathbb Z\to G$ with the property that $\phi(x)=f(x)$ for all $x\in S$. This is just a highbrow formulation for the homomoprhism $n\mapsto f(1)^n$.
There is another one-element set that generates $\mathbb Z$, namely $\{-1\}$. So we have $\mathbb Z=\langle 1\rangle=\langle -1\rangle$. In both cases there is no relation for the genearator $a$. Infact, such a relation could only have the form $a^n=1$ (written multiplicatiuvely!) and that would make $\langle\,a\mid a^n=1\,\rangle=\mathbb Z/n\mathbb Z$ instead of $\mathbb Z$.
The relation that $\mathbb Z$ is abelian is automatic from it being cyclic. Note that the relation would have to be written as $a^na^m=a^ma^n$, but this relation is trivially correct (and hence redundant) because both sides are just $a^{n+m}=a^{m+n}$.