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Let $a,b$ in a group $G$, where $ab=ba$. $|a|=k, |b|=s,$ where $k,s$ are finite

Prove that $|ab|=$ LCM $(|a|,|b|)$

we say $k,s$ in $N$ are co-prime numbers GCD$(k,s)=1$

xrfxlp
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Wento11
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2 Answers2

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If $|a|$ means the order of $a$, then this isn't true. Take for example $G = \mathbb{Z}/6\mathbb{Z}$ with addition and the elements $2$ and $4$ in $G$.

Edit: This is only a counterexample if coprimality of the orders is not assumed, which is now corrected in the OP.

Richard Jensen
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  • But 2 has order 3, and 4 has order 3 and $\gcd(3,3)=3\ne1$ – almagest Sep 24 '19 at 11:47
  • That's my point. According to the OP, we should have $1 = |0| = $lcm$(3,3) = 3$. – Richard Jensen Sep 24 '19 at 11:55
  • ? I thought you were providing a counter-example. But it isn't because you have not complied with the condition that the orders of $a,b$ are coprime. – almagest Sep 24 '19 at 12:01
  • To be fair, I don't think this answer should be downvoted because it was written at the point when condition gcd=1 wasn't written in the OPs question. Question has been edited. – blue Sep 24 '19 at 12:03
  • Aha. I read the OP wrong then. I just thought it was an explanation of what coprime meant. Let me edit to a proof then :). – Richard Jensen Sep 24 '19 at 12:03
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    @blue Ah, I didn't spot that. Don't know how to undo my downvote without upvoting, which I don't want to do before it is corrected! – almagest Sep 24 '19 at 12:06
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We have $(ab)^n=a^nb^n$. Clearly that is 1 if $n=\text{lcm}(k,s)$. We need to show that we cannot have $a^nb^n=1$ for $0<n<\text{lcm}(k,s)$.

Take $n=q_1k+r_1=q_2s+r_2$, where $0\le r_1<k$ and $0\le r_2<s$.

We have $a^nb^n=a^{r_1}b^{r_2}$. Suppose that is 1. Then $1=a^{kr_1}b^{kr_2}=b^{kr_2}$, so $s|kr_2$. But $s,k$ have no common factor, so $s$ must divide $r_2$ and hence $r_2=0$. Similarly, $r_1=0$. So $n>0$ is a multiple of $k$ and a multiple of $s$ and hence at least lcm$(k,s)$. Contradiction.

almagest
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