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# 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?