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I'm hoping to clear up a few uncertainties in my mind about proving that the identity element and inverses of elements in a group are unique.

Suppose we have a group [itex] \left(G, \ast\right)[/itex]. From the group axioms, we know that at least one element [itex]b[/itex] exists in [itex] G[/itex], such that [itex] a \ast b = b \ast a = a \quad \forall \; a\in G [/itex]. Let [itex] b,c \in G [/itex] be any two elements in [itex] G[/itex] satisfying [itex] a \ast b = b \ast a = a [/itex] and [itex] a \ast c = c \ast a = a \quad \forall \; a \in G[/itex]. We have then, that [tex] b= b \ast c = c [/tex] Hence, as [itex]b[/itex] and [itex] c[/itex] are arbitrary (other than satisfying the "identity property" stated above), the only way this can be true is if, in fact, there is only one, unique, identity element.

Is this correct?

Thanks for your time.

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# Conceptual questions on proving identity element of a group is unique

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