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Let $G$ be an abelian group. If $a$ and $b$ are two elements of order $8$ and $10$ respectively, then the order of the element $a^{-1}b$ is

choose the correct option

$a. 80$

$b. 18$

$c. 2$

$d. 40$

My attempt :I think option $a$ is correct

$|a^{-1}|=|a|=8$ .We know that $$\operatorname{lcm}(m,n)=\frac{mn}{\operatorname{gcd}(m,n)}$$

$\implies |a^{-1}b|=\operatorname{lcm}(|a^{-1}|,|b|)\operatorname{gcd}(|a^{-1}|,|b|)=\operatorname{lcm}(8,10)\operatorname{gcd}(8,10)=40.2=80$.

Edit:let $r=|a^{-1}b|\implies r|\operatorname{lcm(m,n)}\implies r=\operatorname{gcd}(m,n)\operatorname{lcm(m,n)}$

Shaun
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jasmine
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    With the exception of $\lvert a^{-1}\rvert=8$, I don't understand your logic. Could you please give the full statement of the theorem by which you are computing the order of the product? – Sassatelli Giulio Nov 24 '23 at 10:23
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    I think the answer is 40, being the lcm. $(a^{-1})^{40}=b^{40}=e$. – Blitzer Nov 24 '23 at 10:24
  • @SassatelliGiulio second isomorphism theorem – jasmine Nov 24 '23 at 10:27
  • @Blitzer $\gcd(8,10)\neq 1$ – jasmine Nov 24 '23 at 10:28
  • @Blitzer You are correct if gcd equals $1$ – jasmine Nov 24 '23 at 10:29
  • @jasmine I asked for the full statement of the theorem you were using, not for its name. Do you mean: "Given a group $G$, $H<G$ and $N\trianglelefteq G$, then $HN<G$ , $N\trianglelefteq HN$, $H\cap N\trianglelefteq H$ and the map $h(H\cap N)\mapsto hN$ is an isomorphism from $H/(H\cap N)$ to $HN/N$"? – Sassatelli Giulio Nov 24 '23 at 10:46
  • okay nice @SassatelliGiulio for example take$ G=\mathbb{Z} $,$H=$ ,$N= $ then $HN=gcd(m,n)$ and $H\cap N=Lcm(m,n)$ – jasmine Nov 24 '23 at 10:49
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    @jasmine In $\Bbb Z_{40},$ there are elements of order $8$ and of order $10$ but no element of order $80.$ – Anne Bauval Nov 24 '23 at 10:53
  • okay @AnneBauval nice example – jasmine Nov 24 '23 at 10:59
  • @SassatelliGiulio let $r=|a^{-1}b|\implies r|lcm(m,n)\implies r=gcd(m,n)lcm(m,n)$ – jasmine Nov 24 '23 at 11:01
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    @jasmine Excuse me, is this a guessing game about what you've done 1 hour ago? You've written "I think" et cetera. The considerations that informed your work are independent on what we're saying right now. We weren't there to tell you: "The Dick Hammerbush theorem says yada yada."; "Ok, then look at this example"; "Oh, but furthermore this and this..." – Sassatelli Giulio Nov 24 '23 at 11:07
  • @SassatelliGiulio Sorry, my apologies – jasmine Nov 24 '23 at 11:10
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    @jasmine The implication $r\mid \operatorname{lcm}(m,n)\Rightarrow r=\operatorname{gcd}(m,n)\operatorname{lcm}(m,n)$ is unwarranted and false. In fact, we could argue that it's hardly ever true. – Sassatelli Giulio Nov 24 '23 at 11:37

2 Answers2

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The order of $a^2$ and $b^2$ are respectively $4$ and $5,$ which are coprime. The order of $(ab)^2$ is therefore $20,$ which is even. This proves that the order of $ab$ is exactly $40.$

Anne Bauval
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Hint: Note that

$$\begin{align} (a^{-1}b)^{40}&=a^{-40}b^{40}\\ &=(a^8)^{-5}(b^{10})^4\\ &=e^{-5}e^4\\ &=e. \end{align}$$

Shaun
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