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. 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.

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 $g^{n}=g^{mq}=1$. 
