Question

If a, b are two distinct arbitrary elements of a group G and H is a subgroup of G then Ha=Hb if and only if ab^-1 €H​

Answer 1

You have søme mistakes. If 1G ∈ H then we cannot de.rive that a∈H and b∈H.

We have that ei ther Hg1=Hg2 or Hg1∩Hg2=∅

(−$\textless$)

If Ha=Hb⇒a=hb for some h∈H.

Thus ab−1=h∈H.

($\textgreater$−)

Let ab−1∈H. Then ab−1=h∈H⇒a=hb=→a

Answer 2

Concept:

If H is subgroup of G then the Identity element for H and G are same.

Given:

a, b are two distinct elements of group G. H is subgroup of G.

Find:

Ha = Hb if and only if $ab^{-1}\in H$

Solution:

Let I be the identity element of group G.

Since H is subgroup G then I also the identity element of H.

Let Ha = Hb

then, $h_1a=h_2b$, for some $h_1,h_2\in H$

$\Rightarrow h_1^{-1}h_1ab^{-1}=h_1^{-1}h_2bb^{-1}$, for $h_1^{-1}\in H, b^{-1}\in G$ Since G and H are groups

$Iab^{-1}=h_3I$, by inverse property and where $h_3=h_1^{-1}h_2\in H$, by closure property of H.

$ab^{-1}=h_3$, by identity property

Since $h_3$ is arbotrary then $ab^-1\in H$.

Conversely let, $ab^{-1}\in H$

$ab^{-1}=h$, for some $h\in H$

$ab^{-1}b=hb$

$aI=hb$, by inverse property

$a=hb$, by identity property

$a\in Hb$, since h is arbotrary

$Ha=Hb$, since $a=Ia\in Ha$ and a is arbotrary

Hence it is proved that if a and b are two distinct arbitrary elements of a group G and H is a subgroup of G then Ha=Hb if and only if $\mathbf{ab^{-}\in H}$.

#SPJ3