Axiom reduction

by math_explorer, Jan 24, 2013, 1:39 AM

In a semigroup with a right identity $1$ and right inverses,

\begin{align*} 1a &= 1a1 \\ &= 1a(a^{-1})((a^{-1})^{-1}) \\ &= 1\cdot 1((a^{-1})^{-1}) \\ &= 1((a^{-1})^{-1}) \\ &= a(a^{-1})((a^{-1})^{-1}) \\ &= a1 \\ &= a \end{align*}

Therefore $a = ((a^{-1})^{-1}) \Longrightarrow (a^{-1})a = 1$
This post has been edited 1 time. Last edited by math_explorer, Mar 26, 2015, 5:35 AM
Reason: align
abstract algebra
algebra

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4 Comments

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darn people always hate me when I give as an exercise to show right inverses and right identities imply left inverses/identities

by dinoboy, Jan 24, 2013, 2:19 AM

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I don't see why the last line holds. Explain?

by phiReKaLk6781, Jan 24, 2013, 2:40 AM

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At this point doing things like this feels a lot like doing unmotivated functional equations to me.

$1a = a$ holds for all $a$. To prevent confusion change variables: $1x = x$ holds for all $x$, and set $x = (a^{-1})^{-1}$ and apply to line 4 and compare to line 7.

by math_explorer, Jan 24, 2013, 2:44 AM

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Here is another proof (which may or may not be equivalent)

Let the identity be $e$. Pick an element $a$. Now:
$a \cdot a^{-1} \cdot (a^{-1})^{-1} = a \cdot e = a$
$a \cdot a^{-1} \cdot (a^{-1})^{-1} = e \cdot (a^{-1})^{-1}$
Thus $a = e \cdot (a^{-1})^{-1}$.
Multiplying by $a^{-1}$ on the left:
$a^{-1} \cdot a = a^{-1} \cdot e \cdot (a^{-1})^{-1} = a^{-1} \cdot (a^{-1})^{-1} = e$. QED.
This post has been edited 1 time. Last edited by dinoboy, Jan 24, 2013, 3:19 AM

by dinoboy, Jan 24, 2013, 3:18 AM

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