Question

IF a and b are arbitrary distinct elements of group G and H is a subgroup of G 1

then aH=bH iff b-1 ​

Answer 1

Answer:

IF a and b are arbitrary distinct elements of group G and H is a subgroup of G 1 then aH=bH iff b-1 ​ it is proved

Step-by-step explanation:

Definition 1. A group is a set $G$ with an operation defined on its elements, such that

a. For all $g, h \in G, g h \in G$

b. An identity element $1 \in G$ exists such that for all $g \in G, 1 g=g$

c. Each $g \in G$ has an inverse $g^{-1} \in G$ such that $g g^{-1}=1$

d. For all $g, h, j \in G, g(h j)=(g h) j$

Definition 2. A subgroup is a group contained inside another group.

Definition 3. For a set$S$, an object $x$, and some operation, you can define a set $S x=\{s x \mid s \in S\}$

Proof 1. $b \in H b=H a .$

Proof 2. $b \in H a$ means $b=h_{1} a$ for some $h_{1} \in H .$ So every element of $H b$ is $h b=h h_{1} a \in H a$ and every element of $H a$ is $h a=h h_{1}^{-1} b \in H b .$