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Say I have an arbitrary group, and a relation inside this group, e.g. $aba^{-1} = b^{-1}$. How can I prove that left-multiplying $a$ on both sides will also give a valid equation?

Thanks

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    There’s nothing to prove: if $aba^{-1}$ and $b^{-1}$ are the same element of the group, then obviously pre-multiplying each of them by the same element will yield the same result.2012-08-24

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No justification is needed. In the formal predicate calculus with equality, the following is a basic axiom.

For any function symbol $f$ (say in two variables, but this doesn't matter), we have the axiom $\forall x\forall x'\forall y\forall y'(((x=x')\land (y=y'))\implies (f(x',y')=f(x,y))).\tag{$1$}$ (There is a similar axiom for relation symbols $R$.)

These axioms enable us to prove formally the fact that you are asking about. However, these properties are so basic that when we are doing ordinary mathematics, we use them without comment, just as you have in the past used without worry the fact that if $x=2$ and $y=3$ then $x+y=2+3$.

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Firstly while defining a group $G$ under a binary operation $*$,did you notice that for any $a$ and $b\in G$,$a*b \in G$ .If $aba^{-1}=b^{-1}$,I do not see why the left hand and right sides would be not be closed under $*$.