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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 ?

H G Sur
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  • Just to clarify, you haven't studied group theory & don't know what a group is? – πr8 Apr 26 '17 at 10:22
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    It means that $G$ is a group with product indicated by $$ (so instead of writing $gh$ we write $gh$). And of course $a,b\in G$ just means that $a,b$ are two elements of $G$ (possibly the same unless you specify otherwise). – lulu Apr 26 '17 at 10:22
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    Note: I assumed your questions concerned notation only. If you don't know what a group is then you need to go study that....don't start with this problem, just go read up on the basics. – lulu Apr 26 '17 at 10:24
  • @lulu Thank you. I have studied binary operations but didn't came across groups. This question was from university entrance not strictly according to my syllabus. With the notating $g*h$ I can now understand the question. – H G Sur Apr 26 '17 at 10:32

3 Answers3

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$(G,*)$ 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 $G$.

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 $e$.

  • How did you know that the binary operation is multiplication ? – H G Sur Apr 26 '17 at 10:37
  • It is not multiplication. It's just an operation –  Apr 26 '17 at 11:36
  • Then how is $b*b^{-1}=1$ ? – H G Sur Apr 26 '17 at 12:17
  • because $b^{-1}$ is just notation for the inverse of $b$, and I denote the identity by $1$. I recommend you read the Wikipedia article on groups, especially the definition part. –  Apr 26 '17 at 12:23
  • Thank you sir, I now get it :) – H G Sur Apr 26 '17 at 12:44
  • But still, it says "For each a in G, there exists an element b in G, commonly denoted $a^{−1}$ (or −a, if the operation is denoted "+"), such that$ a • b = b • a = e$, where e is the identity element." e=1 for multiplication, is it safe to assume e=1 if no hint is given in question ? – H G Sur Apr 26 '17 at 12:46
  • I just denoted $e$ by $1$. The notation is irrelevant yet I'll change my answer to match wikipedia's notation –  Apr 27 '17 at 09:49
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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.

Iti Shree
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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 $*$.

user296113
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