If (G,*) is a group , $a,b\in G $, then $(b^{-1}*a*b)^3 = $

Question

I got this question from a question paper and the options are as follows :

If $(G,*)$ is a group , $a,b\ \in\ G$, then $(b^{-1}*a*b)^3 = $

a) $(b^{-1})^3*a^3*b^3$

b)$b^{-1}*a^3*b$

c)$b^{-1}*a*b^3$

d)$(b^{-1})^3*a*b^3$

Could anyone explain me what does " $(G,*)$ is a group and $a,b\in G$ " means ?

Answer 1

$(G,*)$ is a group

means that $G$ is a set of elements on which a binary operation $*$ is defined, that is, one that satisfies the group axioms.

$a,b\in G$

means $a$ and $b$ are elements in $G$.

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To answer the question in the title, notice

\begin{align} (b^{-1}*a*b)^3&=(b^{-1}*a*b)*(b^{-1}*a*b)*(b^{-1}*a*b)\\ &=b^{-1}*a*b*b^{-1}*a*b*b^{-1}*a*b\\ &=b^{-1}*a*(b*b^{-1})*a*(b*b^{-1})*a*b\\ &=b^{-1}*a*e*a*e*a*b\\ &=b^{-1}*a*a*a*b\\ &=b^{-1}*a^3*b\\ \end{align}

where I denote the identity by $e$.

Answer 2

G is a group on which these operations will be carried on, so you can write: $G= (b^{-1}*a*b)^3;$ $= b^{-1}*a*b*b^{-1}*a*b*b^{-1}*a*b$;

note: $b*b^{-1}$ is 1 so you get $b^{-1}*a^3*b$ as the answer.

Answer 3

More generaly, you can prove by induction that for $n\in\Bbb N$

$$(b^{-1}ab)^n=b^{-1}a^nb$$ and for the inductive step write $$(b^{-1}ab)^{n+1}=(b^{-1}ab)^n(b^{-1}ab)$$ and use the associativity of the law $*$.