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Let $G$ be a finite group such that $3\nmid |G|$ and $a, b \in G$. Prove: If $a^3 = b^3$ then $a = b$.

My thoughts:

If $3 \nmid |G|$, there must be $0 < r < 3$ such that $|G| = 3m + r$.

Now we get:

$$\begin{align} a^3 = b^3 &\implies a^{3m} = b^{3m}\\ &\implies a^{3m+r} = b^{3m}\cdot a^r\\ &\implies e = b^{3m}\cdot a^r\\ &\implies a^{-r} = b^{3m}. \end{align}$$

But that seems useless as I could not go any further. Any hints?

Shaun
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    Since 3 doesn’t divide |G|, the two numbers are actually relatively prime. Hence, 1 is a linear combination with integer coefficients of 3 and |G|. Therefore… – Alon Amit Jan 16 '22 at 22:09
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    Presumably, you mean finite group, and $3\not\mid |G|.$ – Thomas Andrews Jan 16 '22 at 22:09
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    Hint: This works if you replace $3$ with $p$ prime, but it is not always true nonprimes, unless you change “not divisible by” with another condition.. – Thomas Andrews Jan 16 '22 at 22:17

1 Answers1

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Since $3\nmid |G|$, we have that $\gcd(3,|G|)=1$, so, therefore, by Bézout's Theorem, there exist $x,y\in\Bbb Z$ such that $3x+|G|y=1$. Suppose $a^3=b^3$. We have $a^{|G|}=e=b^{|G|}$. Now

$$\begin{align} a&=a^1\\ &=a^{3x+|G|y}\\ &=(a^3)^x(a^{|G|})^y\\ &=(b^3)^x(b^{|G|})^y\\ &=b^{3x+|G|y}\\ &=b^1\\ &=b. \end{align}$$

Shaun
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