If (G, ⋅) is a group such that (ab)-1 = a-1 b-1, ∀ a, b ∈ G, then G is a/an 

A group is said to be abelian if (a*b) = (b*a) ∀a,b ∈G

Since (G, ⋅) is a group

∴ (ab)^-1 = (b^-1a^-1)             (1)

Given: (ab)^–1 = a^–1b^–1       (2)

From (1) and (2)

b-1a-1 = a–1b–1 

Step 4: Multiply both sides by 𝑎  and 𝑏 

To eliminate the inverses and show that 𝑎 𝑏 = 𝑏 𝑎

Multiply both sides on the right by 𝑏 :

𝑏 ^{− 1} 𝑎 ^{− 1} ⋅ 𝑏 = 𝑎 ^{− 1} 𝑏 ^{− 1} ⋅ 𝑏

Since 𝑏 ^{− 1} 𝑏 = 𝑒  (the identity element),

this simplifies to: 𝑏 ^{− 1} 𝑎 ^{− 1} 𝑏 = 𝑎 ^{− 1} 𝑒 = 𝑎 ^{− 1} 

So we get: b^-1a^-1b = a^-1 

Next, multiply both sides of (1) on the right by 𝑎 :

𝑏 ^{− 1} 𝑎 ^{− 1} 𝑏 𝑎 = 𝑎 − 1 𝑎 

Since 𝑎 ^{− 1} 𝑎 = 𝑒

this simplifies to: 𝑏 ^{− 1} 𝑎 ^{− 1} 𝑏 𝑎 = 𝑒

So we get: 𝑏 ^{− 1} ( 𝑎 ^{− 1} 𝑏 ) 𝑎 = 𝑒 

Since  in a group, applying the inverse gives the identity,

we get: ( 𝑎 ^{− 1} 𝑏 ) 𝑎 = 𝑏 

Now, multiply both sides on the left by a:

𝑎 ( 𝑎 ^{− 1} 𝑏 ) 𝑎 = 𝑎 𝑏 

This simplifies to:  eba = ab,

So we get: ba = ab

Thus, we have shown that 𝑎 𝑏 = 𝑏 𝑎

Therefore it is abelian