1. **Problem:** Prove by mathematical induction that for an abelian group $G$ and $a,b \in G$, the equality $$(a * b)^n = a^n * b^n$$ holds for all positive integers $n$. 2. **Formula and rules:** Since $G$ is abelian, the group operation $*$ is commutative: $$a * b = b * a$$ for all $a,b \in G$. 3. **Base case ($n=1$):** $$(a * b)^1 = a * b$$ and $$a^1 * b^1 = a * b$$ So the statement holds for $n=1$. 4. **Inductive hypothesis:** Assume the statement holds for some $n=k$, i.e., $$(a * b)^k = a^k * b^k$$ 5. **Inductive step:** Show it holds for $n=k+1$: \begin{align*} (a * b)^{k+1} &= (a * b)^k * (a * b) \\ &= (a^k * b^k) * (a * b) \quad \text{(by inductive hypothesis)} \\ &= a^k * b^k * a * b \\ &= a^k * a * b^k * b \quad \text{(since $G$ is abelian, we can rearrange)} \\ &= a^{k+1} * b^{k+1} \end{align*} 6. **Conclusion:** By mathematical induction, the equality $$(a * b)^n = a^n * b^n$$ holds for all positive integers $n$ in an abelian group $G$. This completes the proof.