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Let $G:$ be a group.

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

Then which of the following is an order of $b$?

  1. $31$
  2. $32$
  3. $30$
  4. $2$
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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 $8$)

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