Just started group theory. Need help with some basic concepts. I just started learning group theory and I am really insecure with the mathematics. Maybe some of you could take a look at what I have done so far and help me with the questions I don't understand.

Consider the set of complex numbers $$G=\{1,i,-1,-i\}$$
togehter with the usual multiplication "$\cdot$".


*

*Show that $G$ is a group

*Calculate all subgroups of $G$. Which are self-conjugate

*Calculate the factor group of the only non-trivial subgroup. To which group is the factor group equivalent to?

*Consider $\Bbb Z_n=\{0,1,2,3,...,n-1\}$ with operation $*$ defined as $$i*j=i+j \space \text{mod} \space n$$ Prove that this is a group and show that there is an isomorphism between $G$ and $\Bbb Z_4$.



For 1. I checked all the properties (closure property, identity element, inverse element, associative law) and verified that $G$ is a group
For 2. I got two trivial subgroups $N_1=G; N_2=I=\{1\}$ and one non trivial subgroup $\mathcal N=\{1,-1\}$. I need to check which subgroups $\mathcal N$ are self conjugate (normal). I know that a subgroup is self conjugate if $$\forall g \in G, \forall n \in N,gng^{-1} \in N$$
I found that all subgroups are normal but I wasn't sure because it seemed so trivial. Isn't it the case that for all elements of $G$,  $gg^{-1}=1$ and therefore $gng^{-1}$ will always be $\in N$?
For 3. I am not exactly sure I understand what the factor group does but the definition seems to be $G/N=\{ gN:g\in G \}$. So do I just multiply all of the elements of $G$ with all of the elements of $N$? I get $G$ again. Can that be right? 
For 4. I don't get the operation. Does it mean that if I take the element $1$ and $2$. The operation is $1+2 \space \text{mod} \space n$. That doesn't make any sense to me.
 A: Lets take a look at 3. The factor group (or quotient group as I am more used to calling it) can be kinda difficult to grasp. 
We have a group $G$ and a subgroup $N$. For $g\in G$ we write $gN = \{g n : n\in N\}$, this is called the (left) coset of $N$ by $g$ (to get a right coset you do $Ng$, in general $gN \neq Ng$). Lets have a look at an example:
$$ G = \{1,-1,i,-i\},\quad N = \{1,-1\}$$
Check the following: $1N = N$, $-1 N = N$, $i N = -iN = \{i,-i\}$.
More generally when $N$ is a subgroup of $G$ and $n\in N$ you will always get $nN = N$. Indeed the cosets of $G$ by $N$ partition $G$.
So then $G/N$ is the collection of cosets, in our example: $$G/N =\{ gN : g\in G\} =\{1N,-1N, iN,-iN\}=\{ \{1,-1\},\{i,-i\}\} = \{N,iN\}.$$
When $N$ is normal we can define a group operation on the cosets using the operation of $G$ as:$$ (g_1N)(g_2 N) := (g_1g_2)N$$
It would be a good exercise to write out the multiplication table for the operation on $G/N$ for our example, then you just need to find a name for this that you already know.
