The problem presented is about proving that two relations, $\sim_L$ and $\sim_R$, defined on a group $G$ with respect to a subgroup $H$, are equivalence relations, and that the equivalence classes formed correspond to the left and right cosets of $H$ in $G$. Let's break down the steps.

Problem Breakdown:

1. We are given two relations on a group $G$ with a subgroup $H$:

   • $a \sim_L b \iff a^{-1}b \in H$
   • $a \sim_R b \iff ab^{-1} \in H$

2. We need to prove that:

   • $\sim_L$ and $\sim_R$ are equivalence relations.
   • The equivalence classes formed by these relations correspond to the cosets of $H$ in $G$.

Steps to Show These Are Equivalence Relations:

To show that a relation is an equivalence relation, we need to verify three properties for each relation:

• Reflexive: For all $a \in G$, we need $a \sim a$.
• Symmetric: If $a \sim b$, then $b \sim a$.
• Transitive: If $a \sim b$ and $b \sim c$, then $a \sim c$.

For $\sim_L$:

• Reflexivity: We need to show $a \sim_L a$, meaning $a^{-1}a \in H$. This is true because $a^{-1}a = e$ (the identity element), and since $H$ is a subgroup, $e \in H$.

• Symmetry: If $a \sim_L b$, then $a^{-1}b \in H$. Taking the inverse, $(a^{-1}b)^{-1} = b^{-1}a \in H$. Hence, $b \sim_L a$, proving symmetry.

• Transitivity: If $a \sim_L b$ and $b \sim_L c$, then $a^{-1}b \in H$ and $b^{-1}c \in H$. We need to show that $a \sim_L c$, i.e., $a^{-1}c \in H$. Now, $a^{-1}c = a^{-1}b \cdot b^{-1}c \in H \cdot H = H,$ because $H$ is closed under multiplication. Thus, $a \sim_L c$.

Therefore, $\sim_L$ is an equivalence relation.

For $\sim_R$:

• Reflexivity: We need to show $a \sim_R a$, meaning $aa^{-1} \in H$. Since $aa^{-1} = e \in H$, this is true.

• Symmetry: If $a \sim_R b$, then $ab^{-1} \in H$. Taking the inverse, $(ab^{-1})^{-1} = ba^{-1} \in H$. Therefore, $b \sim_R a$, proving symmetry.

• Transitivity: If $a \sim_R b$ and $b \sim_R c$, then $ab^{-1} \in H$ and $bc^{-1} \in H$. We need to show $ac^{-1} \in H$. We have: $ac^{-1} = ab^{-1} \cdot bc^{-1} \in H \cdot H = H,$ since $H$ is closed under multiplication. Thus, $a \sim_R c$.

Therefore, $\sim_R$ is an equivalence relation.

Equivalence Classes:

• The equivalence classes under $\sim_L$ correspond to the left cosets of $H$ in $G$, i.e., the set of all elements $aH = \{ ah \mid h \in H \}$ for a fixed $a \in G$.
• The equivalence classes under $\sim_R$ correspond to the right cosets of $H$ in $G$, i.e., the set of all elements $Ha = \{ ha \mid h \in H \}$ for a fixed $a \in G$.

Thus, we have shown that both relations are equivalence relations and that their equivalence classes correspond to cosets in $G$.

Do you need further clarification or additional details? Here are five questions related to this topic:

1. How can the definition of cosets be generalized for non-normal subgroups?
2. Why is the concept of equivalence relations crucial in abstract algebra?
3. What are some examples of groups where left and right cosets differ?
4. How does the structure of $G$ change if $H$ is a normal subgroup?
5. How do left and right cosets help in understanding quotient groups?

Tip: When working with cosets, always check if the subgroup is normal. If it is, the left and right cosets will coincide, leading to a well-defined quotient group.