The definition of presentation of a group is as follows:
If $G$ is generated by a subset $S$ and there is some collection of relations, say $R_{1}, R_{2}, \dots, R_{m}$(here each $R_{i}$ is an equation in the elements from $S\cup \{ 1 \}$) such that any relation among the elements of $S$ can be deduced from these, we shall call these generators and relations a presentation of $G$ and write
$$
G = \langle\ S \mid R_{1}, R_{2}, \dots R_{m}\ \rangle.
$$
I understand that a group may have many presentations, but I want to know that if a presentation exactly corresponds to a unique group.
Here is my (possible) reasoning:
If we have two groups of the same presentation, then since the generators generate all the elements in group, we can map each product of generators in one group to the other, and since any relations of the generators can be derived from the relations given in the presentation, we can determine when two elements are exactly the same thus the map is well defined and a isomorphism. So a presentation corresponds to a unique group in the isomorphic sense.
If my reasoning is right, then for a given presentation, if I find a group which can have the same presentation, then that group is exactly the group that the presentation corresponding to, is that right?
Best Answer
They are unique up to isomorphism. That is, for groups $G,H$ with a shared presentation $P$, we have
$$\boxed{G\cong H.}$$
For details, see Magnus et al.'s, "Combinatorial Group Theory: Presentations of Groups in Terms of Generators and Relations."
Note: This does not mean that $G,H$ are necessarily equal. For example, the groups $G=(\{-1,1\},\times)$ and $H=\Bbb Z/2\Bbb Z$ are clearly not equal and yet
$$P=\langle a\mid a^2\rangle$$
is a presentation for both.