Prove equality by applying group generator relations on RHS and LHS to get $e=e$? Suppose we have a finite group $G$ with a set of generators $\{a,b,c, \dots \}$ and all the generator relations needed to generate a group.
We want to find if a finite product of generators on left-hand side is equal to another finite product of generators on the right-hand side:
$$ p_{1}(a,b,c,\dots) \stackrel{?}{=} p_2  (a,b,c,\dots)$$
Can we check the truth value of this expression simply by multiplying both sides by generators and their inverses, packing it all up into the identity elements on both sides by using generator relations, so that if:
$$e=e \implies p_1 = p_2$$
$$g=e \implies p_1 \neq p_2$$
Where $g$ is an element of $G$ which is (obviously to us) not the identity element.
If you try something similar on fields, you will sometimes come to a false conclusion, because you can multiply for example both sides by $0$. In groups, there is no equivalent to zero, so my guess is that this can be done, but I don't know how to justify my claim.
A simple example:
The group $\mathbf{C_{3v}}\ = \{ e, C_3, C_3^2, \sigma_v, \sigma_v C_3, \sigma_v C_3^2 \}$ with generator relations $C_3^3=e;\quad \sigma_v^2=e; \quad (\sigma_vC_3)^2=e.$
\begin{align}
\sigma_vC_3 &\stackrel{?}{=} C_3^2\sigma_v \\
\implies \sigma_vC_3 \cdot \sigma_vC_3 &=C_3^2\sigma_v \cdot \sigma_vC_3 \\
\implies (\sigma_vC_3)^2 &= C_3^2 \cdot e \cdot C_3 \\ 
\implies e &= e 
\end{align}
So this would mean that indeed the proposed equality is true.
 A: Yes, that will always work.
First, the case of equality:
Suppose you have an equation $E_1 = E_2$, and you manage to transform this equation into $e = e$ by using only the following operations:

*

*Multiply both sides of the equation by the same expression (resulting in an equation like $E_1 F = E_2 F$ or $F E_1 = F E_2$).

*Replace any sub-expression $F_1$ anywhere in the equation with another expression $F_2$, as long as it is previously known that $F_1 = F_2$. (By "previously known," I mean to imply, in particular, that you can't use $E_1 = E_2$, or any of the other equations you've derived from it, in order to prove that $F_1 = F_2$).

You will end up with a list of equations where the first one is $E_1 = E_2$ and the last one is $e = e$. At this point, you can simply reverse the order of the list, and you'll have a list of equations where the first one is $e = e$ and the last one is $E_1 = E_2$. Furthermore, the truth of each equation in this new list follows from the truth of the preceding equation, because both of the operations above are reversible.
Second, the case of inequality. This case is a little easier. Suppose that, from the equation $E_1 = E_2$, you are able to prove, by any means, that $g = e$, where it is already known that $g \ne e$. This constitutes a proof by contradiction that $E_1 \ne E_2$.
