Does Group Order Not Divisible by 3 Ensure Commutativity if (ab)³ = a³b³?

[Scigatt]

Homework Statement

Let G be a finite group whose order is not divisible by 3. Suppose (ab)³ = a³b³ (\foralla,b\inG)
Prove G must be abelian.

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

Homework Equations

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

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 \forallg \in G. This implies that \forallg \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 θ^z 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.

 

[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.

 

[Scigatt]

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.

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

 

[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

 

[Scigatt]

(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

Ok, thanks