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I've constructed the necessary condition proof for the following, but I am not sure how to prove the sufficient condition:

Show that the group $G$ is abelian if and only if $\forall a,b \in G, \forall n \in \mathbb{N}$ the following holds true: $(ab)^n=a^nb^n$

Proof:

$(\implies)$: Assume that $G$ is an abelian group, we prove that the following holds:

$(ab)^n=a^nb^n$

For $n=2$:

$(ab)^2=(ab)(ab)=abab=aabb=a^2b^2$

For $n=3$:

$(ab^3)=(ab)(ab)(ab)=ababab=aabbab=a^2babb=a^2abbb=a^3b^3$

Assume that

$(ab)^{n-1}=a^{n-1}b^{n-1}$

We prove that $(ab)^n=a^nb^n$

$(ab)^n=(ab)^{n-1}(ab)=a^{n-1}b^{n-1}(ab)=a^{n-1}b^{n-1}ab=a^{n-1}ab^{n-1}b=a^nb^n$

Now obviously in order to prove the sufficient condition we have to assume $G$ is an abelian group and then prove that $(ab)^n=a^nb^n$. But I don't really know how I can use induction to prove this.

user26857
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downmath
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  • I don’t really understand if you want to prove the converse but if you want to, look at your equality for $n = 2$ ! – hugo_panch Jan 16 '22 at 18:57

2 Answers2

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You have already shown that if $G$ is abelian, then $(ab)^n = a^n b^n$. Now to show the implication in the other direction--that is to say, if $(ab)^n = a^n b^n$ for all $a, b \in G$ and $n \in \mathbb N$, then $G$ is abelian, let $n = 2$; then the given condition is that for all $a, b \in G$, $(ab)^2 = a^2 b^2$, and we want to establish that $G$ is abelian; i.e., $ab = ba$ for all $a, b \in \mathbb G$.

To this end, $(ab)^2 = a^2 b^2$ implies $$abab = aabb,$$ and now consider $$a^{-1} abab b^{-1} = a^{-1} aabb b^{-1}.$$

heropup
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  • This is where I'm confused. Do I have to now do the same for, say, $n=3$ and then introduce the induction hypothesis in order to move on with the proof, or is what you have done up until this point sufficient? – downmath Jan 16 '22 at 19:12
  • @downmath I don't see why you want to introduce induction in this part of the proof. Unless I'm missing something, there is no more work to be done. You're trying to prove that the group is abelian, i.e. $ab = ba$ for any $a, b \in G$. The technique above using $n = 2$ clearly shows this to be true, once you clear away the $a^{-1}a$ and $bb^{-1}$ terms. – Novice Jan 16 '22 at 19:41
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Maybe somebody will come along and correct me, but I don't see why induction is necessary here.

For sufficiency, let $n = 2$, so we get

\begin{align*} (ab)^2 &= a^2 b^2\\ \Rightarrow abab &= aabb\\ \Rightarrow a^{-1} abab b^{-1} &= a^{-1} aabb b^{-1}\\ \Rightarrow ba &= ab. \end{align*}

Edit: For the forward direction, couldn't you just do the following:

\begin{align*} (ab)^n &= \underbrace{(ab)(ab)\dots(ab)}_{n\text{ times}}\\ &= abab\cdots ab \text{ (associativity)}\\ &= \underbrace{aa\dots a}_{n\text{ times}}\underbrace{bb\dots b}_{n\text{ times}} \text{ (commutativity)}\\ &= a^n b^n. \end{align*}

Novice
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  • The problem is that $$\underbrace{x\dots x}_{n\text{ times}}$$ is not defined rigorously. – Shaun Jan 16 '22 at 19:30
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    @Shaun Okay, well I suppose the OP handled that part more rigorously than I did. Thanks. – Novice Jan 16 '22 at 19:46