If we let $\phi:\mathbb R^*\to \mathbb R^*$ under multiplication be given by $\phi(x)=$| $x$ |.
What are the left and right cosets of the kernel? Any assistance would be appreciated.
abstract-algebragroup-theory
If we let $\phi:\mathbb R^*\to \mathbb R^*$ under multiplication be given by $\phi(x)=$| $x$ |.
What are the left and right cosets of the kernel? Any assistance would be appreciated.
Best Answer
Kernel is the set of elements which map to identity under $\phi$
$Ker(\phi)=\{x\in\mathbb{R}^*: \phi(x)=1\}$
$Ker(\phi)=\{x\in\mathbb{R}^*: |x|=1\}$
Now which elements in $\mathbb{R}^*$ satisfy this?
Edit (Details):
You have $Ker(\phi)=\{1,-1\}$
Note: In your case, the left and right cosets are equal (Since $\mathbb{R}^*$ is abelian)
You can see it for the finite case below.
Now coming back to your question, let $r\in\mathbb{R}^*$
$$r\cdot Ker(\phi)=Ker(\phi)\cdot r=\{-r,r\}$$