If H is a subgroup of a group G and $a\in G, b\in G,$ then :

Question

Accepted Answer

To determine why Option 4 is the correct answer, we need to analyze the relationship between the cosets $Ha$ and $Hb$ in the context of group theory.

### Explanation for Option 4:

**Option 4: $Ha = Hb \Leftrightarrow ab^{-1} \in H$**

In group theory, the left coset of an element $a$ in a subgroup $H$ of a group $G$ is defined as $Ha = \{ha \mid h \in H\}$. Two cosets $Ha$ and $Hb$ are equal if and only if there exists an element $h \in H$ such that $ha = b$. This can be rewritten as $a = h^{-1}b$, which implies $a^{-1}b = h$. Since $h \in H$, it follows that $a^{-1}b \in H$. By rearranging, we get $ab^{-1} \in H$, which is exactly what Option 4 states. Therefore, Option 4 correctly describes the condition under which the cosets $Ha$ and $Hb$ are equal.

### Explanation for why other options are incorrect:

**Option 1: $Ha = Hb \Leftrightarrow ab \in H$**

This option suggests that the product $ab$ must be in $H$ for the cosets $Ha$ and $Hb$ to be equal. However, the equality of cosets depends on the relative position of $a$ and $b$ with respect to $H$, specifically through the element $ab^{-1}$, not $ab$. Therefore, this condition is not sufficient or necessary for the equality of the cosets.

**Option 2: $Ha = Hb \Leftrightarrow (ab)^{-1} \in H$**

This option implies that the inverse of the product $ab$ must be in $H$. However, similar to Option 1, the equality of cosets is determined by the element $ab^{-1}$, not $(ab)^{-1}$. The condition $(ab)^{-1} \in H$ does not correctly capture the relationship needed for the cosets to be equal.

**Option 3: $Ha = Hb \Leftrightarrow a^{-1}b \in H$**

This option states that $a^{-1}b$ must be in $H$ for the cosets to be equal. While this condition is close to the correct reasoning, it is not the standard form used to express the equality of cosets. The standard form involves $ab^{-1}$, as described in Option 4. Therefore, while $a^{-1}b \in H$ is a related condition, it is not the conventional expression for coset equality.

In conclusion, Option 4 is the correct answer because it accurately describes the condition under which the left cosets $Ha$ and $Hb$ are equal, using the element $ab^{-1}$ in the subgroup $H$.