Is this proof okay? It just seemed too simple
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Prop: If a group G satisfies ab=ca => b=c, then the group is Abelian.
Pf: Assuming ab=ca, then b=c. Thus we can substitute b in for c, and we get: ab=ba. Thus G is Abelian.
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Also, I am supposed to find as many ways to prove (a−1)−1=a as possible.
I found only two..
1:
(a−1)−1a−1=e=aa−1 by inverse propertythus, (a−1)−1=a by RHS cancelation.
2:
(a−1)−1a−1=e=a−1(a−1)−1 since a−1 is the inverse of both a and (a−1)−1 then a must be equal to (a−1)−1 by the uniqueness of the inverse.
Thanks,
-Daon
---------------
Prop: If a group G satisfies ab=ca => b=c, then the group is Abelian.
Pf: Assuming ab=ca, then b=c. Thus we can substitute b in for c, and we get: ab=ba. Thus G is Abelian.
---------------
Also, I am supposed to find as many ways to prove (a−1)−1=a as possible.
I found only two..
1:
(a−1)−1a−1=e=aa−1 by inverse propertythus, (a−1)−1=a by RHS cancelation.
2:
(a−1)−1a−1=e=a−1(a−1)−1 since a−1 is the inverse of both a and (a−1)−1 then a must be equal to (a−1)−1 by the uniqueness of the inverse.
Thanks,
-Daon