After studying homomorphisms from cyclic groups, it is natural to examine functions that send each element of a group to one of its powers. The focus keyword power maps on groups refers to maps such as $\phi(a)=a^n$. These maps are easy to write but not always homomorphisms. In a commutative group, powers of a product split as expected, so the map $a\mapsto a^n$ preserves multiplication. In a noncommutative group, the expression $(ab)^n$ may not equal $a^nb^n$. This lesson separates the reliable abelian case from the dangerous non-abelian case and proves a useful isomorphism criterion for finite commutative groups. :::definition[Power Map on a Group] Let $G$ be a group and let $n$ be an integer. The power map with exponent $n$ is the function $\phi_n:G\to G$ defined by $$\begin{equation*} \phi_n(a)=a^n \end{equation*}$$ for all $a\in G$, whenever the exponent notation is interpreted in the group $G$. ::: The tempting identity $(ab)^n=a^nb^n$ is not true in every group. For $n=2$, it would say $abab=aabb$, which would force a commutation relation between $a$ and $b$ in many cases. In an abelian group, all elements commute, and the formula is valid. Thus power maps on groups should be treated with care: the formula is natural, but the homomorphism property depends on the group operation. :::theorem[Power Map on a Commutative Group] Let $G$ be a commutative group and let $n$ be a positive integer. Define $\phi:G\to G$ by $\phi(a)=a^n$. Then $\phi$ is a homomorphism. ::: :::proof Given that $G$ is a commutative group and $\phi:G\to G$ is defined by $\phi(a)=a^n$. To prove that $\phi$ is a homomorphism. Let $a,b\in G$. Since $G$ is commutative, $$\begin{align*} \phi(ab)&=(ab)^n\\ &=\underbrace{(ab)(ab)\cdots(ab)}_{n\text{ factors}}\\ &=\underbrace{aa\cdots a}_{n\text{ factors}}\underbrace{bb\cdots b}_{n\text{ factors}}\\ &=a^nb^n\\ &=\phi(a)\phi(b). \end{align*}$$ Hence $\phi$ is a homomorphism. $\square$ ::: :::theorem[Finite Commutative Power Isomorphism] Let $G$ be a finite commutative group and let $n$ be a positive integer satisfying $\gcd(n,|G|)=1$. Define $\phi:G\to G$ by $\phi(a)=a^n$. Then $\phi$ is an isomorphism. ::: :::proof Given that $G$ is a finite commutative group, $n$ is a positive integer, $\gcd(n,|G|)=1$, and $\phi(a)=a^n$. To prove that $\phi$ is an isomorphism. Since $G$ is commutative, $\phi$ is a homomorphism. It remains to prove that $\phi$ is one-to-one and onto. Let $a\in\ker\phi$. Then $$\begin{equation*} a^n=e. \end{equation*}$$ Therefore $o(a)\mid n$. By Lagrange's theorem, $o(a)\mid |G|$. Hence $$\begin{equation*} o(a)\mid \gcd(n,|G|). \end{equation*}$$ Since $\gcd(n,|G|)=1$, we get $o(a)=1$. Thus $a=e$, so $\ker\phi=\{e\}$. By the kernel test, $\phi$ is one-to-one. Since $G$ is finite, a one-to-one map from $G$ to itself is onto. Hence $\phi$ is an isomorphism. $\square$ ::: :::example[Squaring on Positive Real Numbers] Let $\phi:(\mathbb{R}^{+},\cdot)\to(\mathbb{R}^{+},\cdot)$ be defined by $\phi(a)=a^2$. Since $\mathbb{R}^{+}$ under multiplication is commutative, $\phi$ is a homomorphism. Indeed, for $a,b\in\mathbb{R}^{+}$, $$\begin{equation*} \phi(ab)=(ab)^2=a^2b^2=\phi(a)\phi(b). \end{equation*}$$ Its kernel is $\{1\}$. With codomain $\mathbb{R}^{+}$, every $y>0$ has $y=(\sqrt{y})^2$, so the map is onto. ::: :::example[Squaring in a Noncommutative Group] In a noncommutative group, the squaring map need not be a homomorphism. In $S_3$, take $a=(12)$ and $b=(23)$. Then $a^2=e$ and $b^2=e$, so $a^2b^2=e$. But $ab=(12)(23)$ is a $3$-cycle, and therefore $(ab)^2\neq e$. Hence $$\begin{equation*} (ab)^2\neq a^2b^2. \end{equation*}$$ Thus the squaring map on $S_3$ is not a homomorphism. ::: :::solved-problem Let $G$ be a finite commutative group of order $15$. Prove that $\phi:G\to G$ defined by $\phi(a)=a^2$ is an isomorphism. ::: :::solution Let $G$ be a finite commutative group with $|G|=15$, and let $\phi:G\to G$ be defined by $\phi(a)=a^2$. To prove that $\phi$ is an isomorphism. Since $G$ is commutative, the power map $\phi(a)=a^2$ is a homomorphism. Also, $$\begin{equation*} \gcd(2,15)=1. \end{equation*}$$ By the finite commutative power isomorphism theorem, $\phi$ is an isomorphism. For a direct kernel check, if $a\in\ker\phi$, then $a^2=e$. Hence $o(a)\mid 2$. Since $o(a)\mid 15$, we get $o(a)\mid 1$. Thus $a=e$, and $\ker\phi=\{e\}$. Therefore $\phi$ is one-to-one. Since $G$ is finite, it is onto. Hence $\phi$ is an isomorphism. $\square$ ::: :::solved-problem Let $G$ be a commutative group. Prove that the inversion map $i:G\to G$ defined by $i(a)=a^{-1}$ is a homomorphism. ::: :::solution Let $G$ be a commutative group, and let $i:G\to G$ be defined by $i(a)=a^{-1}$. To prove that $i$ is a homomorphism. Let $a,b\in G$. Then $$\begin{align*} i(ab)&=(ab)^{-1}\\ &=b^{-1}a^{-1}\\ &=a^{-1}b^{-1} \quad [\because G \text{ is commutative}]\\ &=i(a)i(b). \end{align*}$$ Hence $i$ is a homomorphism. $\square$ ::: :::tip Before using a power map as a homomorphism, identify whether the group is commutative or whether a special argument proves $(ab)^n=a^nb^n$. Without such a reason, the power map on groups may fail the homomorphism condition. ::: :::exercise 1. Prove that $a\mapsto a^5$ is an automorphism of every finite commutative group of order $12$ only if the required coprimality condition holds. Does it hold? 2. Show that $a\mapsto a^3$ is an isomorphism on every finite commutative group of order $10$. 3. In an abelian group, prove that $a\mapsto a^{-2}$ is a homomorphism. ::: :::answer 1. The condition is $\gcd(5,12)=1$, which holds. Therefore the fifth-power map is an isomorphism. 2. Since $\gcd(3,10)=1$, the third-power map is an isomorphism by the theorem. 3. For $a,b\in G$, $(ab)^{-2}=a^{-2}b^{-2}$ because $G$ is abelian. Hence the map is a homomorphism. ::: :::faq[Frequently Asked Questions] Q: Is every power map a homomorphism? A: No. The group operation must allow $(ab)^n=a^nb^n$, which is guaranteed in commutative groups. Q: Why does $\gcd(n,|G|)=1$ matter? A: It forces any element satisfying $a^n=e$ to have order $1$, giving a identity-only kernel. Q: Does one-to-one imply onto for every group? A: No. The finite self-map condition is used in the theorem. ::: :::call-to-action[Continue to Cayley Theorem] subtitle: Represent every abstract group as a group of permutations through a canonical isomorphism. button: Study Cayley Theorem | /abstract-algebra/homomorphisms-and-isomorphisms-of-groups/homomorphisms-of-groups/cayley-theorem-and-permutation-groups button-ghost: Review Cyclic Homomorphisms | /abstract-algebra/homomorphisms-and-isomorphisms-of-groups/homomorphisms-of-groups/homomorphisms-between-cyclic-groups :::