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Let $G$ be a group. If $(ab)^n=a^nb^n$ $\forall a,b \in G$ and $(|G|, n(n-1))=1$ then prove that $G$ is abelian.

What I have proven is that:

If $G$ is a group such that $(ab)^i = a^ib^i$ for three consecutive integers $i$ for all $a, b\in G$, then $G$ is abelian.

A proof of this can be found in the answers to this old question.

user1729
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Ri-Li
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2 Answers2

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We can assume that $n>2$. Since $(ab)^n = a^nb^n$, for all $a,b\in G$, we can write $(ab)^{n+1}$ in two different ways: $$(ab)^{n+1} = a(ba)^nb = ab^na^nb,$$ and $$(ab)^{n+1} = ab(ab)^n = aba^nb^n.$$ Hence, $$ab^na^nb = aba^nb^n.$$ Cancel $ab$ on the left and $b$ on the right to obtain $$b^{n-1}a^n = a^nb^{n-1}.$$ Note that this is true for all $a,b\in G$. (This says that the $n$th power of any element of $G$ commutes with the $(n-1)$st power of any element of $G$.)

Now let $x,y\in G$ be arbitrary; we want to show that $x$ and $y$ commute. Since the order of $G$ is prime to $n$, the $n$th power map $t\mapsto t^n$ on $G$ is bijective, so there exists $a\in G$ such that $x = a^n$. Since the order of $G$ is prime to $n-1$, there exists $b\in G$ for which $y=b^{n-1}$. Therefore, $xy = a^nb^{n-1} = b^{n-1}a^n = yx$. Because $x$ and $y$ were arbitrary, it follows that $G$ is commutative.

ADDED:

Lemma. Let $G$ be a group of finite order $m$, and let $k$ be a positive integer such that $(k,m)=1$. Then the $k$th power map $x\mapsto x^k$ on $G$ is bijective.

Proof. Since $G$ is finite, it suffices to show that the map $x\mapsto x^k$ is surjective. To this end, let $g\in G$; we show that $g$ is the $k$th power of some element in $G$. Since the order of $g$ divides the order of the group $G$, it follows that $(\left| g\right|, k) = 1$. Therefore, $\langle g\rangle = \langle g^k\rangle$. Hence, there is an integer $r$ for which $g = (g^k)^r = g^{kr} = (g^r)^k$. This completes the proof.

James
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  • Why there exists $b\in G$ for which $y=b^{n-1}$? – Ri-Li Sep 14 '14 at 14:14
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    nth power map is automorphism. So it holds but for n-1 th power, what is the reason? – Ri-Li Sep 14 '14 at 14:15
  • @James Is it true in general that the map $g\mapsto g^k$ is bijective whenever $k$ is coprime to the order of the group? As the OP says, in our case it is true for $n$ since we know by assumption that this map is a homomorphism. In general, however, it is not. Could you please explain (or give some reference) why the map is still bijective? – Amitai Yuval Sep 14 '14 at 18:27
  • @AmitaiYuval. I've added an explanation. – James Sep 14 '14 at 19:05
  • @James yes, of course, thanks. – Amitai Yuval Sep 14 '14 at 19:13
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I have got another answer though it can also be easily viewed from James answer too.

We can assume that $n>2$. Since $(ab)^n = a^nb^n$, for all $a,b\in G$. Then we will get $$b^{n-1}a^n = a^nb^{n-1}.$$ this is true for all $a,b\in G$.

Now see Since the order of $G$ is prime to $n$, the $n$th power map $t\mapsto t^n$ on $G$ is automorphism so is the $(1-n)$ th power map $t\mapsto t^{1-n}$ on $G$.

so there exists $a\in G$ & $b\in G$ such that $x = a^n$ & $y=b^{n-1}$. Therefore, $xy = a^nb^{n-1} = b^{n-1}a^n = yx$. Because $x$ and $y$ were arbitrary, it follows that $G$ is commutative.

Ri-Li
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