# Group commutativity problem

1. Oct 9, 2011

### Scigatt

1. The problem statement, all variables and given/known data
Let G be a finite group whose order is not divisible by 3. Suppose (ab)3 = a3b3 ($\forall$a,b$\in$G)
Prove G must be abelian.

Known:
G is a finite group
o(G) not divisible by 3
(ab)3 = a3b3 for all a,b$\in$G

2. Relevant equations
θ:G → G s.t. θ(g) = g3 $\forall$g $\in$ G
Group axioms
Homomorphism properties
Lagrange's theorem
Modular arithmetic?

3. The attempt at a solution
From the facts given above, it's obvious that θ is a homomorphism. Since o(G) is not divisible by 3, then by Lagrange's Theorem, no subgroup of g has order 3. In particular, o(g) ≠ 3 $\forall$g $\in$ G. This implies that $\forall$g $\in$ G a3 = e iff a = e, therefore ker θ = {e}.

Then let a,b $\in$ G s.t. θ(a) = θ(b)
=> θ(ab-1) = θ(a)θ(b-1) = θ(a)θ(b)-1 = e
=> ab-1 $\in$ ker θ
=> ab-1 = e => a = b
Thus θ is 1-1.
Since G is finite, we can use the pigeonhole principle to prove θ is onto. Thus θ is a group automorphism. Since the set of all automorphisms of G is a group, then for any integer z, then θz is an automorphism of G.
If I can prove that either functions i(g) = g-1 or d(g) = g2 are in <θ>, then I'm done, but I don't think that's true in general. Other than that I don't know how to proceed further, though.

Last edited: Oct 9, 2011
2. Oct 10, 2011

### Deveno

this is an interesting problem!

applying θ to aba-1, we get:

ab3a-1 = a3b3a-3
b3 = a2b3a-2
b3a2 = a2b3

but θ is an automorphism (bijective), so every element of G is uniquely a cube.

that is, b3 = c for some unique c in G.

so a2c = ca2, for all a,c in G.

that is, a2b = ba2, for all a,b in G

now because (ab)3 = a3b3:

a(ba)2b = a(a2b2)b, so that

(ba)2 = a2b2.

starting from a2b = ba2, then, we get:

a2b2 = ba2b
(ba)2 = (ba)(ab)
ba = ab.

3. Oct 10, 2011

### Scigatt

How'd you use θ to get ab3a-1 = a3b3a-3?

4. Oct 10, 2011

### Deveno

(aba-1)3 = (aba-1)(aba-1)(aba-1)

= ab(aa-1)b(aa-1)ba-1 = ab(e)b(e)ba-1

= ab3a-1,

that is θ(aba-1) = ab3a-1.

but θ is a homomorphism, θ(aba-1) = θ(a)θ(ba-1)

= θ(a)θ(b)θ(a-1) = a3b3(a-1)3

= a3b3a-3

5. Oct 10, 2011

Ok, thanks