Prove that is group is abelian

Question

Let G be a set of all non-zero real numbers. And a * b = (ab/2).

We need to show (G,*) is an abelian group.

I know I need to show ab = ba, but that seems so trivial,

Assuming the group is abelian,

a * b = b* a

ab/2 = ba/2 Crossing out 2, we get ab = ba

Is this the correct answer ?

Multiplying both sides inverse of a and b results in 1 = 1 . Am I doing something wrong ? Or am I missing something?

Assuming that the group is abelian is not a good start to a proof that the group is abelian. — ancient mathematician, Aug 04 '18 at 06:38
I don't think it's correct to assume the thing you want to prove to be true. — poyea, Aug 04 '18 at 06:39
You're missing the fact that to show something is an Abelian group requires proving that it is a group. — Angina Seng, Aug 04 '18 at 06:43
Then, How should I proceed? The person who taught me this used similar strategy. Can anyone point me in the right direction/ — Kushal Niroula, Aug 04 '18 at 06:44
First attend to group axioms: associativity requires $a(bc)=(ab)c$. Then there's identity and inverses... — Angina Seng, Aug 04 '18 at 06:49

score 0 · Accepted Answer · answered Aug 04 '18 at 06:59

We have that $$a*b=\dfrac{ab}{2}=a\cdot b\cdot\dfrac{1}{2}$$form the other side$$b*a=b\cdot a\cdot\dfrac{1}{2}$$since real numbers with multiplication operator is an Abelian group (If we needed to show that the proof became more complicated) so is this group.

Prove that is group is abelian

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