# Abelian Group

This interesting example is like solving simultaneous equations in Group, using only one Group property tool (Inverse => cancellation law)

Let Group G, ∀a,b ∈G,
for any 3 consecutive integers i,
$(a.b)^{i}= a^{i}.b^{i}$
Prove：G is abelian?
[Herstein: i, i+1,i+2]

Proof:
(a.b)ⁿ= aⁿ.bⁿ …(1)
(a.b)ⁿ⁺¹= aⁿ⁺¹.bⁿ⁺¹ …(2)
(a.b)ⁿ⁺² = aⁿ⁺².bⁿ⁺² …(3)
Take inverse (1):
(a.b)⁻ⁿ = (aⁿ.bⁿ)⁻¹ = b⁻ⁿ.a⁻ⁿ …(4)

Left * (4) to (2)
(a.b)ⁿ⁺¹(a.b)⁻ⁿ =ab

{Right *} (4) to (2):

ab = (aⁿ⁺¹.bⁿ⁺¹).(b⁻ⁿ.a⁻ⁿ) = aⁿ⁺¹.b.a⁻ⁿ
{Left *} a⁻¹
=> a⁻¹(ab) = b = (a⁻¹aⁿ⁺¹).b.a⁻ⁿ = aⁿ.ba⁻ⁿ
Right x aⁿ
=> b.aⁿ = aⁿ.b(a⁻ⁿaⁿ) = aⁿ.b …(5)
Take inverse of (2):
(a.b)⁻ⁿ⁻¹= b⁻ⁿ⁻¹.a⁻ⁿ⁻¹ …(6)
Right x (6) to (3)
(a.b) = aⁿ⁺².bⁿ⁺².(b⁻ⁿ⁻¹.a⁻ⁿ⁻¹)
ab = aⁿ⁺².b.a⁻ⁿ⁻¹ …(7)
Right x aⁿ⁺¹
abaⁿ⁺¹ = aⁿ⁺².b
Left x a⁻¹
baⁿ⁺¹ = aⁿ⁺¹.b …(8)
(b.aⁿ).a = (aⁿ.b).a from (5) b.aⁿ = aⁿ.b
aⁿ.b.a = aⁿ⁺¹.b
cancellation law:
=> ba= ab
=> G is abelian [QED]

Note:

Trick is inverse (1) then x to (2):

b.aⁿ = aⁿ.b …(5)
Similarly inverse (2) then x (3):
baⁿ⁺¹ = aⁿ⁺¹.b …(8)
Solve (5) & (8): by cancellation law
aⁿ.b.a = aⁿ⁺¹.b
=> ba= ab