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asked in Set Theory & Algebra by Veteran (10.2k points)   | 128 views
Is it $31$?
yes, but how???

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Given that, $a^5 = e$ and $aba^{-1} = b^2$, $a,b \in G$

Order of an element of a group is the smallest $m$ such that $a^m = e$. So, $O(a) = 5$

$a^2ba^{-2} = a(aba^{-1})a^{-1} = ab^2a^{-1} = (aba^{-1})(aba^{-1}) = b^2b^2 = b^4$


$\color{navy}{a^3ba^{-3} = a^2(aba^{-1})a^{-2} \\= a^2(b^2)a^{-2} \\= a(aba^{-1})(aba^{-1})a^{-1} \\= a(b^2b^2)a^{-1} \\= (aba^{-1})(aba^{-1})(aba^{-1})(aba^{-1}) \\= b^2b^2b^2b^2 \\= b^8}$


Similarly, $a^4ba^{-4} = b^{16}$ and $a^5ba^{-5} = b^{32}$

  • In a group inverse of identity element is identity element itself. 

So, $a^5 = e$ and $a^{-5} = e^{-1} = e$ 

Multiplying both sides by $b^{-1}$, we get $\Rightarrow$

$b^{31} = e \Rightarrow \color{red}{O(b) = 31}$

Reference : http://fmwww.bc.edu/gross/MT310/hw02ans.pdf (See example $7$ and $8$)

answered by Veteran (24.5k points)  
selected by
thank you, really nice question and explaination.
Yes, it's really a good question.

Really very nice question and nice explanation as well by u @mcjoshi..The step :

ab2a-1=(aba-1)(aba-1) was the trickiest step in my opinion which did not strike to me.

Thanks Habib.
How is this derived: ab^2a−1=(aba^−1)(aba^−1)

Thanks
$a*a^{-1} = e$(identity element)
Sorry, I still am not able to get how $ab^{2}a^{−1} = (a.b.a^{−1})(a.b.a^{−1})$. Thanks
$a*a^{-1} = e$. So, $ab^2a^{-1}$ can be written as : $ab(a^{-1}a)(a^{-1}a)(a^{-1}a)ba^{-1}$, means writing $(a^{-1}a) $anywhere does not change meaning. You can consider $(a^{-1}a) = 1$
Thank you, that was helpful.


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