Left-invertible elements of monoid form submonoid

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Revision as of 22:25, 31 July 2008 by Vipul (talk | contribs) (New page: ==Statement== Let <math>M</math> be a monoid with neutral element <math>e</math>. Then, the set of left-invertible elements of <math>M</math> form a submonoid of <math>M</math>. ==Re...)
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Statement

Let M be a monoid with neutral element e. Then, the set of left-invertible elements of M form a submonoid of M.

Related facts

Inverse map is involutive: Specifically the reversal law for inverses.

Proof

Given: A monoid M with neutral element e

To prove: e is left-invertible, and further, if a_1, a_2 are left-invertible elements of M, so is a_1 * a_2

Proof: Clearly, e is left-invertible, since e * e = e.

Suppose a_1,a_2 are left-invertible and b_1,b_2 are left inverses for a_1, a_2. Then consider:

(b_2 * b_1) * (a_1 * a_2) = ((b_2 * b_1) * a_1) * a_2 = (b_2 * (b_1 * a_1)) * a_2 = (b_2 * e) * a_2 = b_2 * a_2 = e

Thus, b_2 * b_1 is a left inverse for a_1 * a_2, so a_1 * a_2 is left-invertible, completing the proof.