Presentation of a group can be defined in two ways:
Definition 1. A group $G$ is said to have presentation $\langle S \, | \, R\rangle$ if $G$ is isomorphic to $F/N$ where $F$ is the free group on $S$ and $N$ is the conjugate closure of $R$ in $F$.
Definition 2. A group is said to have presentation $\langle S \, | \, R \rangle$ if $G$ is isomorphic to the group $F/\sim$ where $F$ is a free group on $S$ and $\sim $ is an equivalence relation on $F$ and is defined as follows: for all $w_1, w_2 \in F$, $w_1 \sim w_2$ if $w_2$ can be obtained from $w_1$ in finitely many moves of adding or removing $x^{-1}x$ or $xx^{-1}$ for some $x\in S$ anywhere and adding or removing a relator $r \in R$ in the word $w_1$. The group operation on $F/\sim$ is given by $[w_1]\cdot [w_2] =[w_1 w_2]$.
To begin the proof, I define a map $\varphi : F \to F/\sim$ given by $\varphi (w)=[w]$ where $[w]$ denotes the $\sim$ -equivalence class of the word $w$. It is clear that $\varphi$ is an onto homomorphism. If we show that $\ker \varphi = N$ then we will be done. It is easily observed that any word of the form $w_1 r_1 w_1^{-1}w_2 r_2 w_2^{-1}\ldots w_n r_n w_n^{-1}$ where $w_1,w_2,\ldots w_n$ are in $F$ and $r_1 , r_2 , \ldots , r_n$ are in $R$ is in the kernel. So, $N\subseteq \ker f$.
To prove the reverse inclusion, I make the following claim:
Claim. If $w_1 \sim w_2$ and $w_2 \in N$ then $w_1 \in N$.
Proof. Note that it suffices to assume that $w_2$ is of the form $wrw^{-1}$ where $w$ is a word and $r$ is relator and show that a word $w'$ obtained by any single move of adding or remove a relator $r'$ or adding $xx^{-1}$ or $x^{-1}x$ to $w_2$ is also an element of $N$.
We cover these cases separately.
Case 1. Adding a relator $r'$ in $wrw^{-1}$.
- If $r'$ is added to the extreme left and extreme right then we are done.
- If $r'$ is added in between $w$ and $r$, i.e. $wr'rw^{-1}$ then we can write $wr'rw^{-1}=(wr')rr'(wr')^{-1}=((wr')r(wr')^{-1})((wr')r'(wr')^{-1})$. If $r'$ is added in between $r$ and $w^{-1}$, then it can dealt similarly as in the previous case.
- If $r'$ is added in between the word $w$, then after adding $r'$ in $wrw^{-1}$, we have the form $(w_1 r' w_2)r(w_1 w_2)^{-1}$ where $w=w_1w_2$ and $w_1$ and $w_2$ are nonempty words. But then $(w_1 r' w_2)r(w_1 w_2)^{-1} = ((w_1 r' w_1^{-1}))((w_1w_2)r(w_1 w_2)^{-1})$
- If $r'$ is added in between $r$ then after adding $r'$ in $wrw^{-1}$, we have the word $wr_1 r' r_2 w^{-1}$ where $r=r_1 r_2$ where $r_1$ and $r_2$ are nonempty. Since $r_2=r^{-1}$, we have $wr_1 r' r_2 w^{-1}=wr_1 r' (wr_1)^{-1} $.
Case 2. Removing a relator $r'$ in $wrw^{-1}$.
If $r'$ is a subword of $w$ alone, then we have only these cases: $w=r'w_2$ and $w=w_1 r' w_2$ and $w=w_1 r'$. In the first case, we have $wrw^{-1}=r'w_2rw_2^{-1}r'^{-1}$. After removing $r'$, we obtain $w_2rw_2^{-1}r'^{-1}$ which is in $N$. In second case and third case, we obtain $((w_1w_2)r(w_1w_2)^{-1})(w_1r'w_1)^{-1}$ and $w_1rw_1^{-1}r'^{-1}$ respectively and both of which are in $N$.
If $r'$ is a subword of $wr$ then we can write $wr=w_1r'r_2$. Then after removing $r'$ from $wrw^{-1}$, we have $w_1r_2w^{-1}$ which equals $w_1 r_2 r r_2^{-1} r'^{-1} w_1{-1}$ after using that $wr=w_1r'r_2$. Then finally $wrw^{-1}= ((w_1 r_2)r(w_1r_2)^{-1})(w_1r'w_1^{-1})^{-1}$ which is in $N$.
If $r'$ is a subword of $rw^{-1}$, then it can be handled similarly as in part 2.
If $r'$ is a subword of $w^{-1}$ alone then it can be handled similarly as in part 1.
Case 3. Adding $xx^{-1}$ or $x^{-1}x$ to word $wrw^{-1}$ where $x \in S$
Adding $xx^{-1}$ or $x^{-1}x$ makes no difference as it is the empty word.
Case 4. Removing $xx^{-1}$ or $x^{-1}x$ to word $wrw^{-1}$ where $x \in S$
We deal with $xx^{-1}$ alone as dealing with $x^{-1}x$ is no different. Since we demand that words are reduced in $F$, $xx^{-1}$ cannot be a subword of $w$, $r$ or $w^{-1}$. So, we must have the following two cases: "$w=w_1x$ and $r=x^{-1}r_2$" and "$r=r_1x$ and $w^{-1}=x^{-1}w_2$". We deal with the first case alone as the second is similar. So suppose $w=w_1x$ and $r=x^{-1}r_2$. Then after removing $xx^{-1}$ from $wrw^{-1}$, we have $w_1r_2w^{-1}$ which equals $((w_1r_2)^{-1}r(w_1r_2))^{-1}$.
This ends the proof of the claim.
Now, if $w\in \ker f$ then $w\sim \emptyset$ and since $\emptyset \in N$ (because $N$ is a subgroup and hence must contain the identity) so $w$ must be in $N$.
Thus, by the fundamental homomorphism theorem, $F/N$ is isomorphic to $F/\sim$. This proves that the two definitions are equivalent.
Is this proof correct? I'm also looking for alternative proofs, so, those are welcome too.