Let $a,b\in G$ elements of order $5$. If $a^3=b^3$ then $a=b$. [duplicate]

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?

marked as duplicate by Bill Dubuque abstract-algebra StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Jun 13 at 23:43

• $\mathbb Z_{10}$ is not a group under multiplication – J. W. Tanner Feb 15 at 18:46
• What do you mean $o(2)=5$ and $o(2)=8$? – J. W. Tanner Feb 15 at 18:47
• @J.W.Tanner The order of those elements in the group structure. – Randall Feb 15 at 18:48
• How can the order of $2$ be $5$ and $8?!$ – J. W. Tanner Feb 15 at 20:06
• 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. – Gregory Nisbet Feb 15 at 20:27

Since $$b^5=a^5 =a^3a^2 =b^3a^2\implies a^2=b^2$$

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

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

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).

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.$$

• 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. – Bill Dubuque Feb 15 at 19:53

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.