The result is not true if your semigroup does not have an identity. Consider the semigroup \(\{n \in \mathbb{N} : n \ge 2\}\) under ordinary multiplication, which satisfies the cancellation laws but does not have an identity.

The result is not true for infinite unital semigroups. Consider the semigroup \(\mathbb{N}\) under ordinary multiplication.

The result is true for finite unital semigroups. The cancellation laws guarantee that (for example) left multiplication by \(x \in G\) is injective. Why does that guarantee the existence of a right inverse for \(x\)?

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TopNewestThe result is not true if your semigroup does not have an identity. Consider the semigroup \(\{n \in \mathbb{N} : n \ge 2\}\) under ordinary multiplication, which satisfies the cancellation laws but does not have an identity.

The result is not true for infinite unital semigroups. Consider the semigroup \(\mathbb{N}\) under ordinary multiplication.

The result is true for finite unital semigroups. The cancellation laws guarantee that (for example) left multiplication by \(x \in G\) is injective. Why does that guarantee the existence of a right inverse for \(x\)?

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But same problem can we prove in this way by using this statemet that is in a group G a,b,x,y belongs to G ax=b and ya=b have unique solution

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