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M is a commutative monoid with identity element denoted by e.

U(M) is the set of all invertible elements of M.

Let a∈M and b∈M. Prove that if ab∈U(M), then a∈U(M).

I'm not sure if the way I approached this question is legitimate or whether there's a clearer way to show it.

Method:

Let ab ∈ U(M).

Hence $(ab)^{-1}$ ∈ U(M).

⇒ $b^{-1}a^{-1}$ ∈ U(M)

⇒ (ab)($b^{-1}a^{-1}$) ∈ U(M)

⇒ $aa^{-1}$ ∈ U(M)

Similarly for ba ∈ M we have:

$(ba)^{-1}$ ∈ U(M)

⇒ $a^{-1}b^{-1}$ ∈ U(M)

⇒ $a^{-1}b^{-1}$(ba) ∈ U(M)

⇒ $a^{-1}$a ∈ U(M).

As a ∈ M has a right inverse and left inverse in U(M), we can conclude that a ∈ U(M).

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  • $\begingroup$ Writing $a^{-1}$ presupposes $a$ has an inverse. $\endgroup$ – Michael Albanese Jan 9 '16 at 15:58
  • $\begingroup$ Does this mean that the line "(ab)$^{−1}$∈ U(M)." is also incorrect? $\endgroup$ – Joey Wheeler Jan 9 '16 at 16:02
  • $\begingroup$ No, you know $ab \in U(M)$, so it has an inverse, and that inverse is denoted $(ab)^{-1}$. $\endgroup$ – Michael Albanese Jan 9 '16 at 16:05
  • $\begingroup$ I see. In that case how would you find the inverse of ab? Or if that's the wrong approach to this question, how would you go about proving the required result? $\endgroup$ – Joey Wheeler Jan 9 '16 at 22:22
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Suppose that $ab$ has an inverse $c$. Then $(ab)c = 1$, whence $a(bc) = 1$. Since $M$ is commutative, we also have $(bc)a = 1$ and thus $bc$ is an inverse of $a$.

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