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Prove or disprove: let $G$ be a group and $a,b\in G$ elements of order $5$. If $a^3=b^3$ then $a=b$.

I saw the following example which tries to disprove the theorem: $G=\mathbb{Z}_{10}$ and $a=2,b=8$.

I'm not sure about that part, but $o(2)=5$ and $o(2)=8$. I think that $o(2)=\infty$ because there is not $n$ so $2^n\,mod\,10=1$ but I'm not sure.

Does this example disproves the theorem?

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marked as duplicate by Bill Dubuque abstract-algebra Jun 13 at 23:43

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    $\begingroup$ $\mathbb Z_{10}$ is not a group under multiplication $\endgroup$ – J. W. Tanner Feb 15 at 18:46
  • $\begingroup$ What do you mean $o(2)=5$ and $o(2)=8$? $\endgroup$ – J. W. Tanner Feb 15 at 18:47
  • $\begingroup$ @J.W.Tanner The order of those elements in the group structure. $\endgroup$ – Randall Feb 15 at 18:48
  • $\begingroup$ How can the order of $2$ be $5$ and $8?!$ $\endgroup$ – J. W. Tanner Feb 15 at 20:06
  • $\begingroup$ Does this still hold if the order(s) of $a$ and $b$ are merely multiples of five? I have a proof that relies on the fact that $a^3b^2 = b^3a^2 \implies a = b$ that's similar to the others here. $\endgroup$ – Gregory Nisbet Feb 15 at 20:27
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Since $$b^5=a^5 =a^3a^2 =b^3a^2\implies a^2=b^2$$

so $$b^3=a^3 =a^2a =b^2a\implies a=b$$

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  • $\begingroup$ I feel like a moron because I can't follow this. I'm sure it's obvious. Can someone explain the first line? $\endgroup$ – Randall Feb 16 at 4:30
  • $\begingroup$ Never mind, I got it. Hypotheses matter. $\endgroup$ – Randall Feb 16 at 4:33
  • $\begingroup$ @Randall It's just the (subtractive) Euclidean gcd calculation $\, (5,3) = (2,3) = (2,1)\,$ as I explain in may answer (and its comments). $\endgroup$ – Bill Dubuque Feb 16 at 16:22
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If you square both sides, you get $a^6=b^6$. But $a^5=e=b^5$, so $a=b$.

To comment on your solution, note that $o(2)=\infty$ is impossible in a finite group. In $\mathbb{Z}_{10}$ (which always implies the additive structure), the order of $2$ is $5$ (nice and finite).

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No, it is true, the special case $\,m,n = 3,5\,$ below.

Lemma $\ \gcd(m,n) = 1,\ a^m = b^m,\ a^n = b^n\,\Rightarrow\, a = b$

Proof $\ $ The set $\,S\,$ of $\,k\in \Bbb Z\,$ such that $\,a^k = b^k\,$ is closed under subtraction hence, by a 1-line proof, its least positive element $\,d\,$ divides every element, so $\,d\mid m,n\,$ coprime, thus $\,d =1.$

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  • $\begingroup$ For subtraction closure: inverting $2$nd: $\large a^{\large -n} = b^{\large -n}$ times $1$st $\large \,\Rightarrow\, a^{\large m-n} = b^{\large m-n}.\,$ The proof amounts to doing the (subtractive) Euclidean gcd algorithm on the exponents. We could instead use the Bezout GCD identity on the exponents to deduce $\large \,a^d = b^d\,$ for $\large \,d =\gcd(m,n),\ $ which is essentially what some of the other answers do in your special case. $\endgroup$ – Bill Dubuque Feb 15 at 19:53
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As others have shown, $a^3=b^3$ implies $a=b,$ where $a$ and $b$ are group elements with order $5.$ Your example of $2$ and $8$ representing integers modulo $10$ does not disprove this theorem. I'm not sure whether you were talking about the integers modulo $10$ under addition or multiplication (the latter is not a group!), but $2+2+2 \not \equiv 8+8+8 \pmod {10}$ and $2 \times 2 \times 2 \not \equiv 8 \times 8 \times 8 \pmod {10}$, so the hypothesis is not met and it is not a problem that $2 \not \equiv 8;$ the theorem is not disproved.

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