Let $G$ be a cyclic (or not) group of order $n$ and let $k$ be an integer relatively prime to $n$. Prove that the map $x \mapsto x^k$ is surjective.

Question

Let $G$ be a cyclic group of order $n$ and let $k$ be an integer relatively prime to $n$. Prove that the map $x \mapsto x^k$ is surjective. Use Lagrange's Theorem (Exercise 19, Section 1.7) to prove the same is true for any finite group of order $n$. (For such $k$ each element has a $k^\text{th}$ root in $G$. It follows from Cauchy's Thoerem in Section 3.2 that if $k$ is not relatively prime to the order of $G$ then the map $x\mapsto x^k$ is not surjective.)

Claim: The map $x \mapsto x^k$ is surjective.

Scrap Work: Let $f: G \to G$, where $x\mapsto x^k$. Fix any $y\in G$. We want to find a $x$ in domain satisfying $f(x)=y$.

Note that $$y=f(x)=x^k=x^{(1-bn)/a}=(x\cdot x^{-bn})^{1/a}=x^{1/a}$$

Proof: Let $f: G \to G$, where $x\mapsto x^k$. Assume that $G$ is a cyclic group of order $n$. (i.e. $|G|=n$ and $x^n=1$ where $x\in G$.) Suppose that $\gcd(k,n)=1$. (i.e. $ak+bn=1$ for $a,b\in\mathbb{Z}$.)

Fix $y\in G$. Consider $x=y^a$. Note that $y^a\in G$, $x,y\in G$, and $$f(x)=x^k=(y^a)^k=y^{1-bn}=y\cdot (y^n)^{-b}=y.$$ This shows that $f$ is surjective.

Claim: Use Lagrange's Theorem to prove the same is true for any finite group of order $n$. (For such $k$ each element has a $k^\text{th}$ root in $G$. It follows from Cauchy's Thoerem in Section 3.2 that if $k$ is not relatively prime to the order of $G$ then the map $x\mapsto x^k$ is not surjective.)

Proof: Recall Lagrange's Theorem which states **(Lagrange's Theorem)**If $G$ is a finite group and $H$ is a subgroup of $G$, then $|H|$ divides $|G|$.

My question is isn't the proof for the second Claim the same as the first.

Answer 1

The difference is how you get this property: $g^{n}=1$ for any $g\in G$. (You used this property in your last step: $y^{n}=1$.)

For the cyclic group case, $G=\langle t\rangle$. Let $g\in G$. Then $g=t^{m}$. So $g^{n}=t^{mn}=(t^{n})^{m}=1$.

For the finite group case, this is proved as a corollary of Lagrange's Theorem. Let $g\in G$. Then $\langle g\rangle\leq G$. By Lagrange, $|g||n$. Let $m=|g|$. Then $n=mq$ for some $q\in \mathbb{Z}$. So $g^{n}=g^{mq}=1$.

Answer 2

Let us check the map ${\phi(x)=x^k}$, for homomorphism. we can see that ${\phi(xy)=x^k y^k = \phi(x)\cdot \phi(y)}$

• Now let us check injectivity. if ${\phi(y)=e \rightarrow y^k=e}$ since GCD(n,k)=1, and according to Lagrange's theorem, k should divide n, but in our case it is not possible, hence y =e, i.e, kernel of G is trivial , therefore ${\phi}$ is injective.
• Now without using Lagrange's theorem, think of a kernel ${K= \{ x \in G|x^k=e\}}$ if GCD(n,k)=1, then there must be some integers s,t so that ${ks+nt=1}$, now , if , ${x \in K}$ then ${x^1= x^{ks+nt} \rightarrow (x^k)^s \cdot (x^n)^t=e}$, Hence ${K=\{e\}}$, so mapping is one-to-one(injective).
• As group G is finite and it is mapped to itself and injective, this mapping is surjective too. So ${\phi }$ is a isomorphism between ${G \rightarrow G}$. This means there must be ${x \in G}$ that can be represented as ${x^k}$