If $a$ and $b$ are generators of the group $G$ and $ab=ba,$ Prove that $G$ is abelian.

Question

I would like to verify if my proof is rigorous enough and is logically correct.

THE PROOF-

Let$\ z$ be an element of $G.$

This means $z$ can be expressed as $a^{n_1} b^{m_1} a^{n_2} b^{m_2} \cdots a^{n_p} b^{m_p}$. Since $ab=ba$ we can shift each $a$ towards the left until it reaches $a^{n_1}$. This would give us $a^{n_1+n_2\cdots n_p} b^{m_1+m_2 \cdots m_p}$.

Let $x,y∈G$ where $x=a^{n_1+n_2\cdots n_p} b^{m_1+m_2\cdots m_p}$ and $y=a^{r_1+r_2\cdots r_q} b^{s_1+s_2\cdots s_q}$

Then $$xy=a^{n_1+n_2\cdots n_p} b^{m_1+m_2\cdots m_p} a^{r_1+r_2\cdots r_q} b^{s_1+s_2\cdots s_q}$$

Now we can repeat the process of moving every $a$ towards the left to get

$$xy=a^{n_1+n_2\cdots n_p+r_1+r_2\cdots r_q} b^{m_1+m_2\cdots m_p+s_1+s_2\cdots s_q}$$

$yx$ would similarly be equal to $a^{r_1+r_2\cdots r_q+n_1+n_2\cdots n_p} b^{s_1+s_2\cdots s_q+m_1+m_2\cdots m_p}$ which is equal to $a^{n_1+n_2\cdots n_p+r_1+r_2\cdots r_q} b^{m_1+m_2\cdots m_p+s_1+s_2\cdots s_q}$

Since $xy=yx$, $G$ is abelian.

Answer 1

It's fine, except for two details:

1. When you express $z$ as $a^{n_1} b^{m_1} a^{n_2} b^{m_2} \cdots a^{n_p} b^{m_p}$, you should have added that these exponents are integers.
2. The final sentence lacks a quantifier: the conclusion should be that, since $xy=yx$ for each $x$ and each $y$ in $G$, then $G$ is abelian.