[Math] Left & Right Cosets of a Kernel

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 $Ker(\phi)=\{1,-1\}$

Definition of Cosets: If $G$ is a group and $H$ is a subgroup of $G$, then

(i) a left coset of H in G is defined as

$gH=\{gh:h\in H\}$ where $g$ is a fixed element of $G$

i.e. if $H=\{h_1,h_2,\ldots,h_k\}$, then $gH=\{g\cdot h_1,g\cdot h_2,\ldots,g\cdot h_k\}$. Same is the case when $H$ is not finite.

(ii) Similarly, a right coset of H in G is defined as

$Hg=\{hg:h\in H\}$ where $g$ is a fixed element of $G$

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.

Suppose $H=\{h_1,h_2,\ldots,h_k\}$

Let $g\in G$ be arbitrary, then

$gH=\{g\cdot h_1,g\cdot h_2,\ldots,g\cdot h_k\}=\{h_1\cdot g,h_2\cdot g,\ldots,h_k\cdot g\}=Hg$

Now coming back to your question, let $r\in\mathbb{R}^*$

$$r\cdot Ker(\phi)=Ker(\phi)\cdot r=\{-r,r\}$$

So, the set of left cosets of $Ker(\phi)$ in $\mathbb{R}^*$ is $S=\{\{-r,r\}:r\in \mathbb{R^*}\}$

NOTE: This is also the set of right cosets of $Ker(\phi)$ in $\mathbb{R}^*$

Related Question