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How would I go about showing that the pair $(G, *)$ is a group if and only if $G$ is a set and $*$ is an associative binary operation on $G$ such that:

  1. (Left Identity) There exists an element $e \in G$ such that $e * g = g$ for all $g \in G$.
  2. (Left Inverses) For each $g \in G$, there exists an element $g^{-1} \in G$ such that $g^{-1} * g = e$.

Any help would be greatly appreciated, thanks!

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If $G$ is a group, conditions $(1)$ and $(2)$ are directly implied by the axioms. So suppose now that $(1)$ and $(2)$ are true. We need to show existence of right inverses and that the identity also works from the right.

Identity

Note that $$g^{-1} * g = e = e * e = g^{-1} * g * e.$$ By $(2)$ there is $h \in G$ such that $h * g^{-1} = e$. Thus $$h * g^{-1} * g = h * g^{-1} * g * e \text{ }\Longrightarrow \text{ } e* g = e* g * e\text{ }\Longrightarrow\text{ }g = g * e$$ due to $(1)$.

Inverses

We have $$g^{-1} * g * g^{-1} = (g^{-1} * g) * g^{-1} = e * g^{-1} = g^{-1} \text{ }\Longrightarrow \text{ } h * g^{-1} * g * g^{-1} = h * g^{-1}$$$$\Longrightarrow\text{ }e * (g * g^{-1}) = e\text{ }\Longrightarrow \text{ } g * g^{-1} = e.$$

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