11 - Setup for Group Morphisms

Consider the map $\phi :\mathbb{Z}\to {\mathbb{C}}^{\times }$ where $n\mapsto i^n$. Let's verify $\phi$ is a group Homomorphism (or Group Morphism). Take any $n_1,n_2\in \mathbb{Z}$. Notice:
$\begin{array}{rl}\phi (n_1+n_2)& =i^{n_1+n_2}\\ & =i^{n_1}{i}^{n_2}\\ & =\phi (n_1)\phi (n_2)\end{array}$
If we consider the values of $n$ that map to $i^n$:

Value $n$	$\phi (n)$
...	...
-1	$-i$
0	1
1	$i$
2	$-1$
3	$-i$
4	1
...	...
Notice it's very similar to Z over nZ - The Integers Modulo n for $n=4$. So the image of $\phi$ is $\{1,i,-1,-i\}$. In general the formula is:
$
\varphi(n) = \begin
1 & n \equiv 0 \bmod 4 \
i & n \equiv 1 \bmod 4 \
-1 & n \equiv 2 \bmod 4 \
-i & n \equiv 3 \bmod 4
\end
$

Notice here that we've:

• partitioned $\mathbb{Z}$ into four subsets:
  • $C_1={\phi }^{-1}(1)=\{\dots ,-4,0,4,\dots \}$, where here ${\phi }^{-1}$ is the preimage (not the inverse)
  • $C_2={\phi }^{-1}(i)=\{\dots ,-3,1,5,\dots \}$
  • $C_3={\phi }^{-1}(-1)=\{\dots ,-2,2,6,\dots \}$
  • $C_4={\phi }^{-1}(-i)=\{\dots ,-1,3,7,\dots \}$
• ... but only one of these partitions is a Kernel of $\phi$. Namely:
  • $C_1=\{n\in \mathbb{Z}|\phi (n)=1\}=\text{Ker}(\phi )$ since $1$ is the identity in ${\mathbb{C}}^{\times }$. Thus $C_1\le \mathbb{Z}$.
  • The other 3 are not subgroups, but they look very similar to $C_1$, only with some offset in the integers. You can say each of the $C_2,C_3,C_4$ is a "translation" of $C_1$. For example:
    • $C_2=\{\dots ,-3,-1,5,\dots \}=1+C_1$
    • Similarly $C_3=2+C_1$
    • $C_4=3+C_1$.

Here we have a clear bijection between $\{C_1,C_2,C_3,C_4\}\to \{1,i,-1,-i\}$. Here the right is a Group, and since we have a bijection then the left set is also a group, provided we do similar operations from ${\mathbb{C}}^{\times }$. Namely, the operation on $\{C_1,\dots ,C_4\}$ is to "add representatives" to get to the resulting partition, for each $n\in C_k$ being our representative for $C_k$.

Operation	Result
$C_2\star C_3$	$9+10=19\in C_4$ so $C_2\star C_3=C_4$. Doing $1+(-2)$ or any other example yields similar results.
$C_3\star C_3$	$C_1$ since $6+6=12\in C_1$.
...	...
What's the pattern? Notice that:

We've defined an equivalence relation $\sim$ on $\mathbb{Z}$ by:
$a\sim b⟺\phi (a)=\phi (b)⟺a-b\in \text{Ker}(\phi )$
So we'll adjust our notation:
$C_1=[0]=[4]=\dots \text{ or use }{C}_1=\bar{0}=\bar{4}=\dots$
Thus:
$C_2=[1],C_3=[2],C_4=[3]$
Then:
$C_2\star C_3=C_4\to [1]+[2]=[1+2]=[3]$
So instead of using $\star$ we can just use normal integer addition to fill our table.

$\star$	$[0]$	$[1]$	$[2]$	$[3]$
$[0]$	$[0]$	$[1]$	$[2]$	$[3]$
$[1]$	$[1]$	$[2]$	$[3]$	$[0]$
$[2]$	$[2]$	$[3]$	$[0]$	$[1]$
$[3]$	$[3]$	$[0]$	$[1]$	$[2]$
So we claim that:

The General Case

For the more general scenario, say we have some $\phi :G\to H$ is a Homomorphism (or Group Morphism). Then:

• $\text{Ker}(\phi )={\phi }^{-1}(1_H)$ (see Kernel) is a normal subgroup of $G$
• For each $h\in H$ if $h\notin \text{image}(\phi )$ then ${\phi }^{-1}(h)=\mathrm{\varnothing }$. Otherwise then ${\phi }^{-1}(h)=g\cdot \text{Ker}(\phi )=\{gk|k\in \text{Ker}(\phi )\}$. Altnernatively, this set is $\text{Ker}(\phi )\cdot g=\{kg|k\in \text{Ker}(\phi )\}$ for any $g\in G$ with $g\in {\phi }^{-1}(h)$. We call these the cosets of $\text{Ker}(\phi )$.

So in our $C_i$ example we had 4 cosets. Similar to above, there's an equivalence relation on $G$ where:
$g_1\sim g_2⟺\phi (g_1)=\phi (g_2)⟺g_1{g}_2^{-1}\in \text{Ker}(\phi )$
The set of distinct cosets (ie: the set of equivalence classes) is a group under the operation:
$[g_1]\cdot [g_2]=[g_1{g}_2]$
Then this group is isomorphic to the image of $\phi$.

We'll do a lot more of this among many different lemmas.