Question about automorphisms Let $G$ be a finite abelian group and let n be a positive integer relatively prime to $|G|$.
a. Show that the mapping $ϕ(x)=x^n$ is an automorphism of $G$.
b. Show that every $x ∈ G$ has an $n^{th}$ root, i.e., for every $x$ there exists some $y∈G$ such that $y^n=x$.
a.To show that the mapping $ϕ(x)=x^n$ is an automorphism we need to show that $ϕ$ is a homomorphism, and $ϕ$ is one-to-one and onto. Let $ϕ:G->G$ be given by $ϕ(x)=x^n$ and let $x,y∈G$.
First we need show $ϕ$ is a homomorphism, that is $ϕ(xy)=ϕ(x)ϕ(y)$. So, $ϕ(xy)=(xy)^n = (xy)(xy)...(xy)=(xn)(yn)=ϕ(x)ϕ(y)$, since the group is abelian, so $ϕ$ is a homomorphism.
Next, suppose $ϕ(x)=ϕ(y)$, then $e=ϕ(x)ϕ(y^{-1})=ϕ(xy^{-1})=(xy^{-1})^n$, here we see that $ord(xy^{-1})$  divides $n$. Since $xy^{-1}∈G$, it’s order must divide $|G|$(Theorem 10.4), but $n$ and $|G|$ are relatively prime so, $ord(xy^{-1})=1$. Therefore $xy^{-1}=e$ and thus $x=y$. This shows that $ϕ$ is one-to-one. Since $ϕ$ is one-to-one and $G$ is finite, then this implies that $ϕ$ is also onto.
b. I don't know how to word the correlation between part a and part b.
 A: Part (b) is actually equivalent to the fact that $\phi$ is "onto". So you only need to prove (a). The assumption about $n$ being relatively prime to $|G|$ is necessary to prove that $\phi$ is injective. What you actually need to check is that $x^n=1$ implies $x=e$ (the neutral element). Try to use the properties of the rank of an element.
If you will prove that your function is "one to one", then the "onto" part follows immediately form the fact that $G$ is finite.
A: Hint: The following approach does things in not quite the intended order. Let $m=|G|$. Since $m$ and $n$ are relatively prime, by Bézout's Identity there exist integers $s$ and $t$ such that $ms+nt=1$. From this we can conclude that for any group element $a$, we have $a=a^1 =(a^m)^s (a^t)^n=(a^t)^n$. 
A: 
So, my attempt at a proof is: $\varphi(xy)=(xy)^n=x^n*y^n=\varphi(x)\varphi(y)$, so $\varphi$ is a homomorphism.

Good. (Note I've changed your formatting for readability. So these aren't quite direct quotes.)

If $\varphi(a)=\varphi(b)$ then $a^n=b^n \implies a=b$, so $\varphi$ is 1-1

How do you know $a^n = b^n \implies a=b$? This is false for an arbitrary finite abelian group $G$ and an arbitrary $n$. For example, $1^2 = (-1)^2$ is true, but $1 \neq -1$ in the group $\{\pm1\}$ (or in any group of $2N$-th roots of unity. You need to use the fact that $n$ and $|G|$ are relatively prime.

if $y\in G$ then $\varphi(y^-1)=(y^-1)^n=(y^n)^{-1} \in G$ [Check?]

This is unclear, and I think that's partially because of a typo. (Did you mean $\varphi^{-1}(y)$, instead of $\varphi(y^{-1})$?) Basically, you need to show that every element in $G$ is an $n$-th root, i.e., for each $y \in G$ you need to find $x \in G$ such that $x^n = y$. Why can you do this? Again, it depends on the fact that $n$ and $|G|$ are relatively prime. Try to figure out how.

and for (b), I'm not sure how to show that $x\in G$ has an $n$-th root? I think this is where the $n$ being relatively prime to $|G|$ comes into play. That means that $an\neq |G|$ for any integer $a\neq 1$.

Another way to think about this is that two integers $a$ and $b$ are relatively prime (a.k.a. coprime) if their GCD is $1$.

Now how do we see that for every $x$ there exists some $y\in G$ such that $y^n=x$.

As I explained above, this follows from what you proved in part (a).

Edit: Your new proof looks good. The reason you're not seeing part (b) from your proof of (a) is because you proved indirectly that $\varphi$ is a surjection. So for (b), think about what what "$\varphi$ is a surjection" means in terms of the elements of $G$. For all $y \in G$, surjectivity of $\varphi$ implies there exists $x \in G$ with $\varphi(x) = y$. In other words, for every $y \in G$, there exists $x \in G$ such that
$$
y = x^n,
$$
which (by definition) says that $y$ is an $n$-th root of $x$.
A: I am reading "Topics in Algebra 2nd Edition" by I. N. Herstein.
This problem is the same problem as Problem 3 on p.64 in Herstein's book.
I solved Problem 3 as follows:
Since $G$ is abelian, $(xy)^n=x^ny^n$ holds.
So, $\phi$ is a homomorphism.
Let $x\in G$ and $x\neq e$.
Then, $o(x)\mid o(G)$.
Since $o(G)$ and $n$ are relatively prime, $o(x)$ and $n$ are also relatively prime.
From this, $x^n\neq e$ holds.
Why?
If $x^n=e$, then $o(x)\mid n$.
Since $x\neq e$ by assmuption, $o(x)>1$.
So, $o(x)$ and $n$ are not relatively prime.
This is a contradiction.
We proved if $x\neq e$, then $x^n\neq e$.
So, $X=e$ is the only solution of the equation $X^n=e$.
This means $\operatorname{Ker}\phi=\{e\}$.
So, $\phi$ is injective.
Since $G$ is finite, $\phi$ is bijective.
So, $\phi$ is an automorphism of $G$.
Since $\phi$ is surjective,  $X^n=x$ has a solution for any $x\in G$.
