# Thread: What's the correct converse of this problem?

1. ## What's the correct converse of this problem?

Which is the correct converse for the pictured question (so you can see the exact wording)?

Converse One: Take the converse of the statement agb=cgd always implies that ab=cd. That is:

Suppose G is a group with the property that ab=cd always implies agb=cgd. Prove or disprove G is abelian.

This seems like a kind of silly statement, and I see no way to prove it. The only way to make ab=cd is if a=c and b=d or if a=b=c=d.

Converse Two: Take the converse of the question. That is:

Suppose that G is an abelian group. Prove that G has the property that agb=cgd always implies ab=cd.

Which is too easy to be part (b) of this question. If G is abelian and agb=cgd, then we have
agb=cgd
gab=gcd
g-1gab=g-1gcd
ab=cd

Which means that G has that property.
TIA.

2. ## Re: What's the correct converse of this problem?

the wording of the question is vague, and i understand your confusion. however, i favor your "converse two" as the correct one.