Theorem: Let $G$ groups and $S\subset G$ such that $\langle S\rangle =G$. (Here $G=\left\{s_1\ldots, s_n:s_i\in S\cup S^{-1}, n\in\mathbb{N}\right\}$.) Let $\varphi:S\to G$ with $\varphi(s)=s$. By the universal property of free groups there exists a unique homomorphism (in fact, epimorphism) $\varphi:F(S)\to G$ with $$F(S)=\left\{w\in S^{\ast}: w \text{ reduced word} \right\}.$$Then $$G\simeq F(S)/{ker(\varphi)}.$$
Here $\langle \langle S\rangle\rangle=\langle \left\{gsg^{-1}:s\in S\cup S^{-1},\ g\in G\right\}\rangle.$
Let $S\subset G$ and $G=\langle S\rangle.$ Then $\langle S\mid T\rangle $ presentation of $G$ if $G=\langle S\rangle$ and $T\subset \ker\varphi$ and $\langle \langle T\rangle\rangle=\ker\varphi$.
I want prove in a detailed way that $\mathbb{Z}_{3}=\langle a\mid a^3\rangle.$
I have this.
Here $S=\left\{a\right\}.$
First, $a$ must be different from $0$. Because if $a=0$, then $\mathbb{Z}_{3}=\left\{0\right\}$.
If $a=1,$ then $0=1+1+1, 1=1, 2=1+1$.
If $a=2$, then $0=2+2+2, 1=2+2^{-1}, 2=2$.
Therefore, $\mathbb{Z}_{3}=\langle a\rangle$, with $a=1$ or $a=2$.
So . . .
How prove $\ker(\varphi)=\langle\langle \left\{a^3\right\}\rangle\rangle$?
I have this:
$\ker\varphi=\left\{s_1\cdots s_n\in F(S): \varphi(s_1\cdots s_n)=0,\ s_i\in \left\{a\right\}\cup\left\{a^{-1}\right\}, n\in\mathbb{N}\right\}$
$=\left\{s_1\cdots s_n\in F(S): s_1+\cdots +s_n=0,\ s_i\in \left\{a\right\}\cup\left\{a^{-1}\right\}, n\in\mathbb{N}\right\}$
In order to show $\ker(\varphi)=\langle \langle a^3\rangle\rangle$, one must first be clear on what $\varphi$ is, and instead of writing, say, $\color{red}{a=1}$, one writes $\varphi(a)=1$.
You are on the right track in that you have evidence to suggest that $\varphi: a\mapsto 1\text{ or } 2$; that is, that the generator $a$ of the presentation $\langle a\mid a^3\rangle$ maps via $\varphi$ to one of the elements of $\Bbb Z_3$ given by a number coprime to $3$.
Ask yourself,
But can you deduce what each word maps to under $\varphi$ in general?
Hover your cursor over (or, if you're on a touchscreen device, tap) the box below for some hints.
I hope this helps :)