$\langle a,b\mid ab=1\rangle\cong \mathbb{Z}$ I'm new to free groups and presentation of groups, and I am having some problems with some basic facts:
Let $\langle a,b\mid ab=1\rangle$ be a group presentation. I want to show that $\langle a,b\mid ab=1\rangle$ is isomorphic to $\mathbb{Z}$. So, by definition of presentation, $\langle a,b\mid ab=1\rangle$ could be understood as the greater free group generated by $\langle a,b\rangle$ where $ab=1$, or equivalently, $b=a^{-1}$. So, let's take an element in the free group generated by $\langle a,b\rangle$, named $x$. By definition of free group, $x$ is a product of consecutive powers of $a$ and $b$, with negative or positive exponents (for example, $a^{2}b^{-2}a^{3}b^{2}a^{-5}$ is an element of $\langle a,b\rangle$). As $b=a^{-1}$, $x$ can be expressed as the product of a bunch of powers of $a$, with entire exponent. Then, if $x\in\langle a,b\mid ab=1\rangle$, $x\in \langle a\rangle$, where $\langle a\rangle$ is the free group with one generator, $a$. Is clear that $\langle a\rangle\subset \langle a,b\mid ab=1\rangle$, so
$$
\langle a,b\mid ab=1\rangle=\langle a\rangle. 
$$
And $\langle a\rangle$ is isomorphic to $\mathbb{Z}$ by definition of free group.
Is this argument well-thought? Thanks in advanced!
 A: There are a few issues with this proof, but the main one is this. You treat $\langle a \rangle$ as a subset of $\langle a, b \mid ab = 1 \rangle$, but it is not. To illustrate this point with a more extreme example, we cannot say that $\langle a \rangle$ is a subset of $\langle a \mid a = 1\rangle$; the former has infinitely many elements, while the latter has but one.
You have thus far only managed to show that $a$ generates the group $\langle a, b \mid ab = 1\rangle$. In other words, you have shown that the “inclusion” map $f : \langle a \rangle \to \langle a, b \mid ab = 1\rangle$, defined as the unique group homomorphism such that $f(a) = a$, is surjective. But you haven’t shown that this “inclusion” is injective.
Here is a more complete proof using the universal property of group presentations. Let $f : \langle a \rangle \to \langle a, b \mid ab = 1\rangle$ be as above, and let $g : \langle a, b \mid ab = 1\rangle \to \langle a \rangle$ be the unique group homomorphism such that $g(a) = a$ and $g(b) = a^{-1}$. Then we see that $f \circ g$ is the unique group homomorphism sending $a$ to $a$ and $b$ to $b$, so it is the identity map. Similarly, $g \circ f$ is the unique group homomorphism sending $a$ to $a$, so $g \circ f$ is the identity map. Thus, $f$ and $g$ are inverse isomorphisms.
A: As in the first two comments,  I get $\langle a,b\mid ab\rangle =\langle a,a^{-1}\mid \rangle =\langle a\mid \rangle \cong \Bbb Z$, because $b=a^{-1}$.
Without the relator $ab$, we get $\langle a,b\mid\rangle =\Bbb Z*\Bbb Z=F_2$, the free group on two generators.   In this group $ab$ has infinite order.
