Does Cross Cancellation Ensure a Group is Abelian?

[BustedBreaks]

I have two proofs that I am uneasy about and one I'm having trouble with so hopefully I can figure out where I'm going wrong if I am. Ignore the weird numbers, its to help me organize the problems.

14) Let G be a group with the following property: Whenever a, b and c belong to G and ab=ca, then b=c. Prove that G is Abelian. ("cross cancellation" implies commutativity.)

[tex]a,b,c \in G[/tex]

[tex]ab=ca[/tex] implies [tex]b=c[/tex]

[tex]ab=ca[/tex]

[tex]b=c[/tex]
replace c with b in [tex]ab=ca[/tex] and

[tex]ab=ba[/tex]

which implies an Abelian groupThe problem I have with this is that I used cancellation and from some reason I get the feeling I'm not supposed to due to the parenthesis. The question seems to answer itself wit that statement because what they give in the question seems to imply cancellation. Unless Abelian means more than just commutative..
26) Prove that if [tex](ab)^{2}=a^{2}b^{2}[/tex] in a group G, then ab=ba.

Here's what I have:
[tex](ab)^{2}=a^{2}b^{2}[/tex]

[tex]a^{2}b^{2}=a^{2}b^{2}[/tex]

[tex]aabb=aabb[/tex]

[tex]a=a[/tex]

Implies either:
[tex]ab=ab[/tex] or [tex]ba=ba[/tex]

[tex]a=\frac{ab}{b}[/tex] [tex]a=\frac{ba}{b}[/tex]

[tex]\frac{ab}{b}=\frac{ba}{b}[/tex]

[tex]b\frac{ab}{b}=b\frac{ba}{b}[/tex]

[tex]ab=ba[/tex]15) Let a and b be elements of an Abelian group and let n be any integer. Show that [tex](ab)^{n}=a^{n}b^{n}[/tex]. Is this also true for non-abelian groups?

I feel like I can use a bit from the above example, by starting off from
here:
[tex](ab)^{n}=a^{n}b^{n}[/tex]

[tex]\overbrace{ab\cdot\cdot\cdot ab}^{n}=\overbrace{a\cdot\cdot\cdot a}^{n}\overbrace{b\cdot\cdot\cdot b}^{n}[/tex]

but I feel like I'm working backwards from this to showing it is abelian instead of going from an abelian group to this which seems like what I should be doing...

 

[Mark44]

I have two proofs that I am uneasy about and one I'm having trouble with so hopefully I can figure out where I'm going wrong if I am. Ignore the weird numbers, its to help me organize the problems.

14) Let G be a group with the following property: Whenever a, b and c belong to G and ab=ca, then b=c. Prove that G is Abelian. ("cross cancellation" implies commutativity.)

[tex]a,b,c \in G[/tex]

[tex]ab=ca[/tex] implies [tex]b=c[/tex]

[tex]ab=ca[/tex]

[tex]b=c[/tex]
replace c with b in [tex]ab=ca[/tex] and

[tex]ab=ba[/tex]

which implies an Abelian group


The problem I have with this is that I used cancellation and from some reason I get the feeling I'm not supposed to due to the parenthesis.

What parenthesis? I don't see any parentheses in the above.

The question seems to answer itself wit that statement because what they give in the question seems to imply cancellation. Unless Abelian means more than just commutative..



26) Prove that if [tex](ab)^{2}=a^{2}b^{2}[/tex] in a group G, then ab=ba.

Here's what I have:
[tex](ab)^{2}=a^{2}b^{2}[/tex]

[tex]a^{2}b^{2}=a^{2}b^{2}[/tex]

(ab)² means (ab)(ab). So you have (ab)(ab) = a²b². a and b are elements of a group, so each has an inverse, right?

[tex]aabb=aabb[/tex]

The two lines below aren't worth saying. It's obviously true that a equals itself, and that ab and ba equal themselves.

[tex]a=a[/tex]

Implies either:
[tex]ab=ab[/tex] or [tex]ba=ba[/tex]

[tex]a=\frac{ab}{b}[/tex] [tex]a=\frac{ba}{b}[/tex]

[tex]\frac{ab}{b}=\frac{ba}{b}[/tex]

[tex]b\frac{ab}{b}=b\frac{ba}{b}[/tex]

[tex]ab=ba[/tex]


15) Let a and b be elements of an Abelian group and let n be any integer. Show that [tex](ab)^{n}=a^{n}b^{n}[/tex]. Is this also true for non-abelian groups?

I think the approach you need to take is an induction proof. You proof with ... is more akin to "armwaving."

I feel like I can use a bit from the above example, by starting off from
here:
[tex](ab)^{n}=a^{2}b^{2}[/tex]

[tex]\overbrace{ab\cdot\cdot\cdot ab}^{n}=\overbrace{a\cdot\cdot\cdot a}^{n}\overbrace{b\cdot\cdot\cdot b}^{n}[/tex]

but I feel like I'm working backwards from this to showing it is abelian instead of going from an abelian group to this which seems like what I should be doing...