# commutativity of a group

• Jan 28th 2010, 12:02 PM
nataliemarie
commutativity of a group
I'm trying to prove this problem but I'm not sure whether or not my answer is working. The problem states

"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."

I have solved for b and c using inverses, and got that b=a-1(ca) and c=(ab)a-1. The second half of the problem states that b=c. So I wanted to use replacement to show that the group is abelian. But I'm not sure if that would prove G is abelian for any arbitrary b and c or if its only commutative for this exact pair.
• Jan 28th 2010, 01:00 PM
clic-clac
Hi

Perhaps it is better to start with two elements, let's say \$\displaystyle a,b,\$ in \$\displaystyle G.\$ You want to prove \$\displaystyle ab=ba\$.

Inverses are indeed often useful for group equations: take a look at \$\displaystyle ab=ab(a^{-1}a)\$ using associativity and your hypothesis.
• Jan 28th 2010, 01:31 PM
nataliemarie
I'm sorry I don't think I understand. I could regroup by the associative property stating
ab=(aba-1)a
and I previously found that c=(aba-1) so by replacement ab=ca so therefore b=c. But I'm still confused about how to prove the group is abelian. I'm missing something.
• Jan 28th 2010, 11:30 PM
clic-clac
Well, you have \$\displaystyle ab=(aba^{-1})a\$

You can define \$\displaystyle c:=aba^{-1}\$ if you want, so, by hypothesis, you have: \$\displaystyle b=c,\$ i.e. \$\displaystyle b=aba^{-1},\$ and you can conclude.
• Jan 29th 2010, 08:32 AM
nataliemarie
yea i realized that a couple minutes after i had replied. (Headbang) thank you sooo much for the explanation!