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The exercise in the book is:

If we have $k$ in the group $(G, \star)$ for three consecutive integers $$\forall a,b \in G \quad (a \star b)^k = a^k \star b^ k$$ Prove that $(G, \star)$ is commutative.

In the book I could only find the prove for $(a \star b)^{-1} = b^{-1} \star a^{-1}$.

I'm trying to solve this exercise and here's what I've done so far:

$( \Rightarrow)$ If $(G,\star)$ is commutative then $\forall a,b \in G$ we have $$a \star b = b \star a$$ So $(a\star b)^k = (b \star a)^k = a^k \star b^k$. Because:

$\underbrace{(a \star b)\star (a \star b) \star ... \star (a\star b)}_k = a^k \star b^k$.

$(\Leftarrow)$ Now if $(a \star b)^k = a^k \star b^k$ then if $a=b$ we have $$(a \star a)^k= a^k \star a^k$$ This implies that $a^k = (\underbrace{a \star ... \star a}_k)= a^k$ Then $\forall a,b \in G$ we have $$(a \star b)^k = a^k \star b^k = (\underbrace{a \star ... \star a}_k) \star (\underbrace{b \star ... \star b}_k)$$ we can show that $b^k \star a^k$. So it's commutative.

I'll be happy if you tell me I'm in the true way or not.

Amirreza Hashemi
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    You've assumed what you want to prove. You're supposed to prove if some $k$ satisfies $(ab)^k=a^kb^k,,(ab)^{k+1}=a^{k+1}b^{k+1},,(ab)^{k+2}=a^{k+2}b^{k+2}$ (where I've suppressed the $\star$s) then $ab=ba$. Hint: first use the first two equations to show $(ab)^ka=a(ab)^k$. – J.G. Jan 14 '23 at 11:29
  • For correct proofs see the duplicates. You should carefully compare your tryings with these answers. – Dietrich Burde Jan 14 '23 at 12:07
  • Your statement is of the form $P\implies Q$ it is not of the form $P\iff Q$. Your start with $P$ and end with $Q$. – A. P. Jan 14 '23 at 12:08

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