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I was working on this question and was unable to find a way to prove it. This is the solution given by the professor. I am not satisfied with the solution. It does not seem like something I or any of my peers would come up with out of nowhere, and I was wondering if anyone had a better idea for proving this that is more satisfying.

Thanks

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MathDoer2320
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  • That depends entirely on what you know... in general, for any abelian group, the result holds, so the fact that you are being asked to show it for a cyclic group suggests you don’t have too many tools on hand. – Arturo Magidin May 13 '20 at 20:51
  • The proof given by your professor is a very standard and straightforward proof. You may want to sit down and read it carefully to see the basic ideas being used. – Anurag A May 13 '20 at 20:51
  • Why do a and b equal what they equal, on the second line? – MathDoer2320 May 13 '20 at 20:53
  • This comes from the fact that if $|g|=n$, then the order of $g^t$ is given by $\frac{n}{\gcd(n,t)}$. Ask yourself, if $|g|=15$, then what would be the order of $g^6$. Say $|g^6|=k$, then $g^{6k}=e$. But then $15 | 6k$. So what should be the smallest positive integer $k$ for this to happen? – Anurag A May 13 '20 at 20:57
  • Then why is n in the numerator (and same for in the 5th line)? – MathDoer2320 May 13 '20 at 20:59
  • Thank you by the way I did not have that in my notes – MathDoer2320 May 13 '20 at 20:59
  • @MathDoer2320 Sorry I had a typo. I fixed it. – Anurag A May 13 '20 at 21:00
  • Ah, thank you.. – MathDoer2320 May 13 '20 at 21:00
  • The trick of expressing $\gcd(a,b)$ as a linear combination of $a$ and $b$ is the key here. It's a pretty ubiquitous and useful trick so it's always worth considering when there's a gcd in the picture. –  May 13 '20 at 21:33

2 Answers2

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By hypothesis, $\exists x,y \in G$ s.t. $|x|=a$, $|y|=b$ and $gcd(a,b)=1$.

Claim: $|xy| = ab$

$proof:$

Since $G$ is cyclic, it is abelian, and so $(xy)^{ab}=x^{ab}y^{ab}=(x^a)^b(y^b)^a=(e^b)(e^a)=(e)(e)=e$.

Therefore $ab$ divides $|xy|$.

Suppose that $\exists k \in \mathbb{N}$ such that $1<k<ab$ and $(xy)^k=e.$ Since $G$ is abelian we have $(xy)^k=x^ky^k=e$. Thus $x^k=y^{-k} \in \langle x \rangle \cap \langle y \rangle = \{e\}$ and so we have that $a|k$ and $b|k$, which then implies $k \geq lcm(a,b)=ab$, but this is a contradiction as we chose $k$ so that $1<k<ab$.

Thus we have showed that $(xy)^{ab}=e$ and that there is no number $1<k<ab$ such that $(xy)^k=e$. Thus the order of $xy$ is $ab$

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    You need to say why $a|k$ and$b|k$ in the second part of the proof. For instance, $x^ky^k=e\implies x^k=y^{-k}\in\langle x\rangle\cap\langle y\rangle=e$ etc. –  May 13 '20 at 23:59
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As @Arturo said, this is true for any abelian group, and more generally it's true for any two elements of a group that commute. A nice proof is here https://math.stackexchange.com/a/371107.