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Let $G$ be a group and $a,b\in G$. Prove that $|a|=|a^{-1}|$ and that $|ab|=|ba|$.

I said that $|a|=n$ where $n$ is the smallest integer such that $a^n=e$ and $|a^{-1}|=m$ where $m$ is the smallest integer such that $(a^{-1})^{m}=a^{-m}=e$ but I don't know how to continue. Any suggest are appreciated.

Tryncha
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2 Answers2

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Hint. By definition, $a^{-n}a^n=e$, where $e$ is the identity element. But $a^n=e$, which gives you one direction. Can you show the other direction? For the second question, consider $(ab)^na=a(bab\dotsm aba)$. What happens if $(ba)^n=e$?

YiFan Tey
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    If I say that $a^{-m}=e$ then multiplicating both equalities I get that $a^{n}a^{-m}=e$ and I conclude that $a^{-m}$ is the inverse of $a^{n}$. So $m=n$? And is this enough to proof the first part? For the second part, if a say that $(ba)^{n}=e$ then $a(ba)^{n}=a(bab\cdots aba)$. And again, is this enough to proof that $m=n$? – Tryncha Oct 06 '19 at 23:06
  • @Tryncha no, it's not sufficient. You have to show two directions: first that the order of $a$ divides $n$ (equivalently $a^n=e$), but also that $n$ divides the order of $a$. The argument above is the first part, I'll leave you to think about the second. – YiFan Tey Oct 06 '19 at 23:09
  • Can the downvoter explain? – YiFan Tey Oct 06 '19 at 23:10
  • I don't know what happened, I just clicked to upvote and it displays -1. And I think I can't change that. Sorry, but your answer helped me. – Tryncha Oct 06 '19 at 23:19
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$$ a^n=e\implies (a^{-1})^n=a^{-n}=(a^n)^{-1}=e^{-1}=e$$

$$(ab)^k=e\implies a(ba)^{k-1}b=e\implies $$

$$( ba)^{k-1}=a^{-1}b^{-1} \implies$$

$$(ba)^k=e$$

Shaun
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Mohammad Riazi-Kermani
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