Prove G is Abelian

Question

Let $G$ be a group and $a,b\in G$.

given that $(ab)^k=a^k b^k$ and $(ab)^{k+2}=a^{k+2} b^{k+2}$ for some $k\in \mathbb N$. prove that $G$ is abelian.

So far my attempt was: $(ab)^{k+2}=(ab)(ab)^k(ab) \Longrightarrow a^{k+2}b^{k+2}=(ab)a^kb^k(ab) \Longrightarrow a^kb^k=b^{-1}a^{k+1}b^{k+1}a^{-1}$

and from here I'm stuck. I noticed there was a similar question but there was given that also $(ab)^{k+1}=a^{k+1}b^{k+1}$ which is missing in my question.

thanks for your help.

What you want to prove is false in general. Let $k = 1$, and $G$ be the group found in this answer. Then of course $(ab)^1 = a^1 b^1 = ab$, and since the group has exponent three it also holds that $(ab)^3 = a^3 b^3 (= e)$. But the group isn't abelian. It's however true for some values of $k$, eg. $k=3$. Where did you find this question? — Najib Idrissi, Jul 13 '15 at 09:30
Which textbook is it? The classical exercise (found in Herstein) is to prove that the groups is abelian when $(ab)^k=a^k b^k$ holds for three consecutive values of $k$. See for instance http://math.stackexchange.com/questions/40996/prove-that-if-abi-aibi-forall-a-b-in-g-for-three-consecutive-integers. — lhf, Jul 13 '15 at 10:13
@lhf I saw this question. I wrote it in my question. It's not a famous textbook so it's not gonna help if I will tell you — Vegetal605, Jul 13 '15 at 10:39

score 2 · Accepted Answer · edited Apr 13 '17 at 12:58

The statement you want to prove is false in general. Let $k=1$ for example, and let $$G = \langle x,y,z \mid x^3 = y^3 = z^3 = 1, yz = zyx, xy = yx, xz = zx\rangle$$ be the group found in this answer. Then $G$ is not abelian, but $G$ has exponent three. It thus holds that $(ab)^3 = e = a^3 b^3$ for all $a,b \in G$, and moreover it obviously holds that $(ab)^1 = ab = a^1 b^1$. There are some $k$ for which the statement is true though, for example when $k = 3$, the group is necessarily abelian. This answer gives more insight on what values of $k$ could make it work.

Prove G is Abelian

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