1. **Problem statement:** Show that for a group $G$ and fixed element $a \in G$, the set $$H_a = \{x \in G : xa = ax\}$$ is a subgroup of $G$. 2. **Recall subgroup criteria:** To prove $H_a$ is a subgroup, we must show: - The identity element $e$ of $G$ is in $H_a$. - $H_a$ is closed under the group operation. - $H_a$ is closed under taking inverses. 3. **Step 1: Identity element in $H_a$** Since $e$ is the identity in $G$, for any $a \in G$: $$ea = a = ae$$ Thus, $e \in H_a$. 4. **Step 2: Closure under group operation** Take any $x, y \in H_a$. By definition: $$xa = ax \quad \text{and} \quad ya = ay$$ We want to show $(xy)a = a(xy)$: \begin{align*} (xy)a &= x(ya) \\ &= x(ay) \quad \text{(since } y \in H_a) \\ &= (xa)y \\ &= (ax)y \quad \text{(since } x \in H_a) \\ &= a(xy) \end{align*} Hence, $xy \in H_a$. 5. **Step 3: Closure under inverses** Take any $x \in H_a$, so $xa = ax$. We want to show $x^{-1} a = a x^{-1}$. Multiply both sides of $xa = ax$ on the left by $x^{-1}$: $$x^{-1} (x a) = x^{-1} (a x)$$ Simplify left side: $$ (x^{-1} x) a = x^{-1} a x$$ $$ e a = x^{-1} a x$$ $$ a = x^{-1} a x$$ Multiply both sides on the right by $x^{-1}$: $$ a x^{-1} = x^{-1} a$$ Rearranged: $$ x^{-1} a = a x^{-1}$$ Thus, $x^{-1} \in H_a$. 6. **Conclusion:** Since $H_a$ contains the identity, is closed under the group operation, and closed under inverses, $H_a$ is a subgroup of $G$. **Final answer:** $H_a$ is a subgroup of $G$.