Skip to content

抽象代数代写| Cosets of Kernels

1. Cosets of Kernels

In this section we will relate two notions in group theory – homomorphisms and kernels – through the partition of a group into cosets of the kernel.

Let

$$
f: G \rightarrow H
$$

be a homomorphism of groups (recall that it means that for any $x, y \in G$, we have $f(x y)=f(x) f(y)$. Consider also the kernel of $f$ :

$$
\operatorname{Ker}(f)=\left{x \in G \mid f(x)=e_{H}\right}
$$

Since $\operatorname{Ker}(f) \subset G$ is a subgroup, we can consider the partition of $G$ into (left) cosets, which are, by definition, subsets

$$
g \operatorname{Ker}(f)={g x \mid x \in \operatorname{Ker}(f)}
$$

2. Question 1

What are the left cosets of $\operatorname{Ker}(f)$ ?

3. Proposition 1

Elements $x, y \in G$ belong to the same left coset of $\operatorname{Ker}(f)$ if and only if

$$
f(x)=f(y)
$$

Proof. Elements $x$ and $y$ belong to the same left coset of Ker $(f)$ if and only if one can multiply $x$ from the left by an element $k$ in $\operatorname{Ker}(f)$ and obtain $y:$

$$
x k=y
$$

This happens if and only if $y^{-1} x \in \operatorname{Ker}(f)$.

By definition, the latter holds if and only if

$$
f(y-1 x)=e_{H}
$$

Using the defining property of homomorphisms, we find that it happens if and only if

$$
f(y)^{-1} f(x)=e_{H} \Longleftrightarrow f(x)=f(y)
$$

Since at each step of our proof we used if and only if statements, we have proved the proposition in both directions.

4. Corollary 1

Left cosets of $\operatorname{Ker}(f) \subset G$ are in 1 -to- 1 correspondence with the image $\operatorname{Im}(f) \subset H .$

5. Remark 1

The entire discussion above holds verbatim for the right cosets.

6. Definition 1: L

$\mathrm{t} f: X \rightarrow Y$ be a map between sets. A fiber over an element $y \in Y$ is the set of all elements in $X$ mapped to this $y$ :

$$
f^{-1}(y)={x \in X \mid f(x)=y} \subset X
$$

Proposition 1 can be now stated as follows: left cosets of $\operatorname{Ker}(f)$ are precisely the nonempty fibers of $f$.

7. Example 1

Consider the sign homomorphism:

$$
\text { sgn: } S_{n} \rightarrow{+1,-1}
$$

Its kernel is the alternating group $A_{n}$. This group has two cosets:

$$
\left.A_{n}={\text { even permutations }}, \quad \text { {odd permutation }\right}
$$

which correspond respectively to elements $+1$ and $-1$ of group ${+1,-1} .$

8. Example 2

Consider a homomorphism

$$
f: \mathbb{Z}{8} \rightarrow \mathbb{Z}{4}
$$

defined by

$$
f(x)=2 x \quad \bmod 4
$$

Elementwise this map can be presented as follows:

$$
\underbrace{\begin{array}{l}
0,2,4,6 \
1,3,5,7
\end{array} \mapsto \underbrace \begin{array} { l }
{ 0 } \
{ 1 } \
{ 2 } \
{ 3 }
\end{array}}{Z{8}}
$$

Thus we see that both fibers $f^{-1}(0)$ and $f^{-1}(2)$ have 4 elements (the size of Ker $\left.(f)\right)$, while the fibers $f^{-1}(1)$ and $f^{-1}(3)$ are empty, because 1 and 3 are not in the image of $f .$

There is an important enumerative corollary of the above proposition.

9. Proposition 2

If $f: G \rightarrow H$ is a homomorphism, then

  • $[G: \operatorname{Ker}(f)]=|\operatorname{Im}(f)|$
  • $|G|=|\operatorname{Im}(f)| \cdot|\operatorname{Ker}(f)|$

Proof. From Proposition 1 we know that the left cosets of $\operatorname{Ker}(f)$ in 1 -to- 1 correspondence with $\mid$ Im $(f) \mid$. This proofs the first claim.

To prove the second claim, we invoke Lagrange’s theorem, which states that for any subgroup $K \subset G$

$$
|G|=[G: K] \cdot|K|
$$

apply it to $K=\operatorname{Ker}(f)$, and substitute $[G: \operatorname{Ker}(f)]$ from the first part.

This proposition can be useful in studying homomorphisms between different groups.

10. Example 3

Consider a homomorphism $f: S_{5} \rightarrow \mathbb{Z}{7} .$ On one hand, by Lagrange’s theorem the size of subgroup $\operatorname{Im}(f) \subset \mathbb{Z}{7}$ divides $7 .$

On the other hand, by the proposition

$$
\underbrace{\left|S_{5}\right|}_{5 !=120}=|\operatorname{Ker}(f) | \operatorname{Im}(f)|,
$$

thus $|\operatorname{Im}(f)|$ is a factor of 120 .

The only option this leaves for $|\operatorname{Im}(f)|$ is 1 , hence $f$ is a trivial homomorphism, i.e., it maps every element of $S_{5}$ to the identity $0 \in \mathbb{Z}_{7}$.

发表回复

您的电子邮箱地址不会被公开。