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Let $G$ be a group with operation $\cdot$ and let $a \in G$. Define a new operation $*$ on the set $G$ by $x*y$ = $x·a·y$ for all $x,y \in G$. Show $G$ is a group under the operation $*$.

Does this group under the operation have an inverse?

I know you need to use $x*b = a^{-1}$ and $b*x = a^{-1}$. But I am not getting them to have the same result.

So far I have
$b*x = bax$ so $bax = a^{-1}$
and
$x*b = xab$ so $xab = a^{-1}$.

rschwieb
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Michelley
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  • What do you get, if you first solve $x$ from the equation $bax=a^{-1}$, and then from the equation $xab=a^{-1}$? Do you get the same answer? – Jyrki Lahtonen Jan 31 '14 at 07:36
  • Yes. Then G under the operation * has an inverse. – Michelley Jan 31 '14 at 07:41
  • Ok, so have checked all the group axioms now? – Jyrki Lahtonen Jan 31 '14 at 07:47
  • "Does this group under the operation have an inverse?" is a strange question. It probably should read: does every element of $G$ have an inverse for the operation "$$", and the answer must be "yes" since that is part of showing that $(G,{})$ is a group. – Marc van Leeuwen Jan 31 '14 at 08:01

2 Answers2

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Hint: look at $x=a^{-1}b^{-1}a^{-1}$

user68061
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If possible, let x' be the inverse(left as well as right) of x in G. then by def we have x*x'=x'*x=e, e=1 be the identity in G. Now, x*x'=e =>xax'=1, where a∈G =>x'=x−1a−1 Similarly, x'*x=e =>x'ax=1, where a∈G =>x'=x−1a−1 Hence for a element in G its inverse exists in G.

BPAL
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