We have a monoid with the identiy $e$, and we know that any element of the monoid squared is the identity. I have to prove now that this is a commutative group.
May somebody help me or give me a hint how I can show that this must be true?
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Math Lover
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RM777
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To show it's a commutative group, you need to show two things:
- Existence of inverses
- Commutativity
For inverses, take any element $a$; $a^2 = e$, so $a$ is its own inverse.
For commutativity, you need to show that for any elements $a,b$, $ab = ba$. Since $a^{-1}=a$ and $b^{-1}=b$, $$ab(ba)^{-1} = aba^{-1}b^{-1}=abab = (ab)^2 = e$$ Thus (multiplying both sides by $ba$ on the right) $ab = ba$
BallBoy
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Can you explain me why (ab)² = e implies ab = ba? – RM777 May 16 '18 at 19:49
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@RM777 I've edited; is it clearer now? – BallBoy May 16 '18 at 19:50
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Yes, appreciate it. – RM777 May 16 '18 at 19:52
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The idea of the proof is that we assume that (ba)^-1 is the inversed element of ab, right? – RM777 May 16 '18 at 20:00
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@RM777 We don't assume it; we prove it. – BallBoy May 16 '18 at 20:03
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But wee took advantage of the proposition that there is only one inverse for each element. – RM777 May 16 '18 at 20:10
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@RM777 Yes. It is a theorem of group theory that inverses are unique. (Actually the proof can be modified so it does not rely on this fact -- first show $ab$ is an inverse to $ba$, then multiply both sides of $abab=e$ by $ba$.) – BallBoy May 16 '18 at 21:05
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$a^2=e$ implies that every $a$ has an inverse, itself. Therefore, this monoid is in fact a group. It is well known that $a^{2}=e$ in a group implies that it is commutative.
lhf
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I know that a² = e implies a group. But if I have two elements in the group which are not the same, say a and b. How can i prove that ab = ba with the fact that a² = e? – RM777 May 16 '18 at 19:41
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"It is well known that $a^2=e$ in a group implies that it is commutative." Not so well-known that the op knows it. That probably should be proven. (Although it'd be fair to ask the OP to figure it out.) – fleablood May 16 '18 at 19:43
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... $e = (ab)(ab) = (aa)(bb)$ so $a(ab)(ab)b = a(aa)(bb)b$ so $ba = ab$. – fleablood May 16 '18 at 19:48
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To prove that $a^2 = e$ for all $a$ implies $ab = ba$:
$(ab)(ab) = e$
$a(ab)(ab)b = a*e*b$
$(aa)(ba)(bb) = ab$
$e*ba*e = ab$
$ba = ab$.
or
$ab = k$
$a=abb = kb$
$e = aa = kba$
$k = ke = kkba = ba$
So $ab = k = ba$
fleablood
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