Let $G=\{-1,1,i,-i\}$ be the fourth roots of unity. Prove $G$ is a group under multiplication and $G$ is isomorphic to $\Bbb{Z}_4$.. 
A) Prove that $G=\{-1,1,i,-i\}$ is a group under multiplication.

a.1) First I need to show that G is indeed closed under the operation *
we have $1*1=1$ where $1 \in G$ 
we have $-1*-1 = 1$ where $1 \in G$
we have $1*-1 = -1$ where $-1 \in G$ and $-1 * 1 =-1 \in G$
we have $1*i=i$ and $i*1 = i$ where $i \in G$
we have $-1*i=-i$ and $i*-1=-i$ where $-i \in G$
let $k \in \mathbb{N}$ then $i^{2k}=-1$ where  $-1 \in G$
Finally let $k \in \mathbb{N}$ then we have $i^{2k+1}=-i$ where $-i \in G$
Okay so I have shown that all possible outcomes from every combination of multiplication between any elements yields an element in $G$ 
a.2) To show that this is a group it must be associative. We may assume that since the roots of unity is associative that $G$ is therefore associative.
a.3) Since $1 \in G$ we can see that:
$i*1=i,\space -i*1=-i,\space 1*1=1, \space -1*1=-1$ This shows that every element has an identity element
a.4) Does every element have an inverse?
 well we have $1*1=e, -1*-1=e$, $i*i*-1 = e , -i*-i=e$ Yes
Which means this is a group. Does a.1-a.4 properly show that G is a group under multiplication and do I have any errors in parts a.1-a.4?

Prove that G is isomorphic to $\mathbb{Z}_4$ under addition

To do this I believe that I must show that $G \to G'$ $($where $G' = \mathbb{Z}_4)$is a bijective homomorphism
To start out we define $\mathbb{\phi}:G \to G'$
I need to show that $\mathbb{\phi}(a)=\mathbb{\phi}(b)$ implies that $a = b$ which proves that its one to one. 
I have no idea how to go about this. I cannot grasp the idea of connecting Roots of unity and modular arithmetic let alone show that they are one to one, onto, and that $\mathbb{\phi}(a,b)=\mathbb{\phi}(a)\mathbb{\phi}(b)$

Part C) Prove that there can be no isomorphism $\varphi:G\to\Bbb Z_4$ such that $\varphi(i)=2.$

I think I can go about this by first assuming as in the previous question that this is under addition. This is false because the addition of any two imaginary yields an imaginary is an imaginary number and can therefore never be equal to 2. Is it that easy or am I missing the whole idea?
 A: A) It is correct, except for associativity. What does “the roots of unity is associative” mean? Anyway, you are working with the standard multiplication in $\mathbb C$, which is associative.
B) Just define $\phi\colon\mathbb Z_4\longrightarrow G$ by $\phi(k)=i^k$ ($k\in\{0,1,2,3\}$). It's a bijection and $\phi(ab)=\phi(a)\phi(b)$.
C) If $\varphi(i)=2$, then $\varphi(i^2)=2\times2=0$, $\varphi(i^3)=3\times2=2$ and $\varphi(i^4)=4\times2=0$. So, $\varphi$ will not be a bijection.
A: Here's a few ideas that could help you out.
It is far easier to write it out differently and use properties that relate the roots to each other. Let's $\xi_4=i$. Then we note that the fourth roots of unity can be written as $\xi_4,\xi_4^2\,\xi_4^3,\xi_4^4$. 
How do we show that the group is closed under multiplication? Well we simply check that $\xi_4^a\xi_4^b=\xi_4^{ab}$ where we can reduce $ab$ modulo $4$ as $\xi_4^4=1$ is contained in the group. It clearly is. Then it's also easy to see that each element has an inverse as well, so it is a group (it clearly is associative).
There's a very clear isomorphism here. What should you map to $x$ if $Z_4=\langle x\rangle$?
A: 
Does a.1-a.4 properly show that G is a group under multiplication [. . .]?

Yes.
For the rest of the question, consider this
Hint: Map $i\mapsto [1]_4$, where $[1]_4\in\Bbb Z_4$. Use the fact that $i$ has order four, as does $[1]_4:=\{a\in\Bbb Z: 4\mid 1-a\}$.
A: If you can recognize that $G$ is a finite subset of the group $(\mathbb{C}\setminus\{0\},*)$, then to prove the point A) you just need of a.1).
For the point B), note that both groups have 1 element of order 2 ($-1$ and $2$, respectively) and 2 elements of order 4 ($i$ and $-i$, and $1$ and $3$, respectively), and that homomorphisms preserve elements' order. So, there are two isomorphisms $\phi\colon \mathbb{Z}_4 \to G$, namely:
\begin{alignat}{1}
&0\mapsto 1 \\
&1\mapsto i \\
&2\mapsto -1 \\
&3\mapsto -i \\
\end{alignat}
(concisely, $\phi(k)=i^k$, clearly a homomorphism) and:
\begin{alignat}{1}
&0\mapsto 1 \\
&1\mapsto -i \\
&2\mapsto -1 \\
&3\mapsto i \\
\end{alignat}
(concisely, $\phi(k)=i^{-k}$, clearly a homomorphism as well).
This actually answers also to the point C), since none of the two isomorphisms maps (via $\phi^{-1}$) $i$ to $2$ (not surprisingly, indeed, since homomorphisms preserve elements' order and $i$ and $2$ have order $4$ and $2$, respectively).
