(adsbygoogle = window.adsbygoogle || []).push({}); 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}= a^{3}b^{3}([itex]\forall[/itex]a,b[itex]\in[/itex]G)

Prove G must be abelian.

Known:

G is a finite group

o(G) not divisible by 3

(ab)^{3}= a^{3}b^{3}for all a,b[itex]\in[/itex]G

2. Relevant equations

θ:G → G s.t. θ(g) = g^{3}[itex]\forall[/itex]g [itex]\in[/itex] 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 [itex]\forall[/itex]g [itex]\in[/itex] G. This implies that [itex]\forall[/itex]g [itex]\in[/itex] G a^{3}= e iff a = e, therefore ker θ = {e}.

Then let a,b [itex]\in[/itex] G s.t. θ(a) = θ(b)

=> θ(ab^{-1}) = θ(a)θ(b^{-1}) = θ(a)θ(b)^{-1}= e

=> ab^{-1}[itex]\in[/itex] 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) = g^{2}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.

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# Group commutativity problem

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