Some simple proofs relating to groups.

[daon]

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 \(\displaystyle (a^{-1})^{-1} = a\) as possible.

I found only two..

1:
\(\displaystyle (a^{-1})^{-1}a^{-1} = e = aa^{-1} \text{ by inverse property} \\ \text{thus, } (a^{-1})^{-1} = a \text{ by RHS cancelation.}\)

2:
\(\displaystyle (a^{-1})^{-1}a^{-1} = e = a^{-1}(a^{-1})^{-1}\\ \text{ since } a^{-1} \text{ is the inverse of both } a \text{ and } (a^{-1})^{-1} \\ \text{ then } a \text{ must be equal to } (a^{-1})^{-1} \text{ by the uniqueness of the inverse.}\)

Thanks,
-Daon

 

[pka]

I think that you have assumed too much.
You must prove that for any a & b in the group, ab=ba.

So say ab=x then right multiply by a: aba=xa.
But we are given that a(ba)=xa implies ba=x or ba=ab.

The other two are fine.

 

[daon]

I'm wondering, why have I assumed too much? Is it because we need to create the situation in which to use the hypothesis? If so, why cant we say right off the bat that since (ab)a = a(ba), ab=ba?

 

[pka]

Clearly written that way, it will work.
But that was not clear to me what you meant.