Every group has a presentation

Question

My question is pretty simple: prove that every group has a presentation. This is always assumed when I read about presentations but no one has actually proved this (as far is I can search the internet). It seems very logical to me that this is true. I've tried saying that if $G$ is a group, then $\langle G\mid\phantom{G} \rangle$ is a presentation for it, but I'm not sure wether this is true.

Answer 1

Every group $G$ has a presentation, yes, since $G$ is spanned by $G$ and by all relations $g.h=gh$, with $g,h\in G$.

Answer 2

We can consider $A\subset G$ a set of generators of $G$ (that could be $G$). Now there is always a free group of words in $A$ that I called $FA$.

By the inclusion map $i: A\to G$ you can apply the universal property of free groups to get an homomorphism $\phi: FA\to G$. This map is surjective because for construction $A$ is a set of generators of $G$ and so, by first homomorphism theorem, you have that

$G\cong \frac{FA}{\ker (\phi)}$

And so a presentation of $G$ is $\langle A \mid \ker(\phi)\rangle$