Skip to content

抽象代数代写| Homomorphisms & Kernels

1. Homomorphisms

2. Definition 1: Homomorphism

Let $(G, *)$ and $(H, \cdot)$ be two groups. A map

$$
f: G \rightarrow H
$$

is called a homomorphism if for any elements $x, y \in G$ we have

$$
f(x * y)=f(x) \cdot f(y)
$$

If $f$ is also bijective, we call it an isomorphism.

3. Example 1

  • Let $S_{n}$ be a permutation group, and $H={+1,-1}$ be a group with respect to multiplication. Then sign of permutation defines a homomorphism

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

Indeed, sgn satisfies the key property:

$$
\operatorname{sgn}(\sigma \tau)=\operatorname{sgn}(\sigma) \operatorname{sgn}(\tau)
$$

  • Consider $G=\mathbb{Z}{6}$ and $H=\mathbb{Z}{3} .$ Define a map

$$
f: \mathbb{Z}{6} \rightarrow \mathbb{Z}{3}
$$

as follows. Take any integer $x$ representing a congruence class in $\mathbb{Z}{6} .$ Then $a=(x$ mod 3$)$ represents a class in $\mathbb{Z}{3}$. This gives a well-defined map $\mathbb{Z}{6} \rightarrow \mathbb{Z}{3}$ :

$$
f(x)=\left(\begin{array}{lll}
x & \bmod 3
\end{array}\right) \in \mathbb{Z}_{3}
$$

This map is an isomorphism, since

$$
(x+y \quad \bmod 3)=\left(\begin{array}{lll}
x & \bmod 3
\end{array}\right)+(y \bmod 3) .
$$

  • For any groups $(G, *)$ and $(H, \cdot)$ there is a trivial homomorphism:

$$
f(x)=e_{H}
$$

for all $x \in G$, where $e_{H} \in H$ is the identity.

From now on, by default, we will use multiplication as the group operation.

4. Proposition 1

Let $f: G \rightarrow H$ be a homomorphism. Then

  1. $f\left(e_{G}\right)=e_{H}$, where $e_{G} \in G$ and $e_{H} \in H$ are identities;
  2. $f\left(x^{-1}\right)=f(x)^{-1}$, for any $x \in G$;
  3. $\operatorname{Im}(f) \subset H$ is a subgroup.

Proof. 1. Let $a=f\left(e_{G}\right) .$ By homomorphism property, we have

$$
a=f\left(e_{G}\right)=f\left(e_{G} e_{G}\right)=f\left(e_{G}\right) f\left(e_{G}\right)=a^{2} .
$$

Thus in $H$ we have $a=a^{2}$, which implies $a=e_{H}$. 2. To prove that $f\left(x^{-1}\right)$ is the inverse of $f(x) \in H$, we need to compute the product:

$$
f\left(x^{-1}\right) f(x)=f\left(x^{-1} x\right)=f\left(e_{G}\right)=e_{H} .
$$

Thus $f\left(x^{-1}\right)$ is the inverse of $f(x)$.

  1. We need to verify two claims:
  • Claim 1: for any $a, b \in \operatorname{Im}(f)$ we have $a b \in \operatorname{Im}(f)$. Indeed we know that there exists $x, y \in G$ such that

$$
f(x)=a \quad f(y)=b .
$$

Thus $f(x y)=f(x) f(y)=a b$ implying that $a b \in \operatorname{Im}(f)$.

  • Claim 2: for any $a \in \operatorname{Im}(f)$ we have $a^{-1} \in \operatorname{Im}(f)$. Indeed, if $a=f(x)$, then by part 2 of this proposition, $a^{-1}=f\left(x^{-1}\right)$.

5. Definition 2: Kernel of a homomorphism

Let $f: G \rightarrow H$ be a homomorphism. A kernel of $f$

$$
\operatorname{Ker}(f) \subset G
$$

is the set

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

6. Proposition 2

Kernel $\operatorname{Ker}(f) \subset G$ is a subgroup. Furthermore, it is a normal subgroup, meaning that for any $x \in \mathrm{Ker}(f)$ and any $g \in G$ we have

$$
g^{-1} x g \in \operatorname{Ker}(f)
$$

Proof. Let $x, y \in \operatorname{Ker}(f)$ be any two elements, i.e.,

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

Then

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

and

$$
f\left(x^{-1}\right)=f(x)^{-1}=e_{H} .
$$

This proves that $\operatorname{Ker}(f) \subset G$ is a subgroup. Now let us prove that $\operatorname{Ker}(f) \subset G$ is a normal subgroup. Indeed for any $x \in \operatorname{Ker}(f)$ and any $g \in G$ we have

$$
f\left(g^{-1} x g\right)=f\left(g^{-1}\right) \underbrace{f(x)}{e{H}} f(g)=f\left(g^{-1}\right) f(g)=f(\underbrace{g^{-1} g}{e{G}})=e_{H} .
$$

Thus by definition $f^{-1} x g \in \operatorname{Ker}(f)$

Problem 1: Give an example of a group $G$ and its subgroup $H \subset G$ which is not normal.

7. Example 2: Alternating group

Let sgn: $S_{n} \rightarrow{+1,-1}$ be the sign homomorphism. Then $\operatorname{Ker}(\operatorname{sgn}) \subset S_{n}$ is the set of all even permutations, and by the above it forms a normal subgroup of $S_{n}$.

This group has a special name: alternating group ans is often denoted as

$$
A_{n}:=\operatorname{Ker}(\operatorname{sgn})
$$

8. Proposition 3

$\left|A_{n}\right|=n ! / 2$

Proof. Since $\left|S_{n}\right|=n !$ and $S_{n}={$ even permutations $} \cup{o d d$ permutations $}$, it is enough to prove that $\mid{$ even permutations $}|=|{$ odd permutations $} \mid$.

To this end we construct a bijection

$$
f=: \text { {even permutations }} \rightarrow \text { {odd permutations }} .
$$

We first define $F: S_{n} \rightarrow S_{n}$

$$
F(\sigma)=(1,2) \sigma
$$

Since every transposition is odd, for $\sigma$ even, $F(\sigma)$ is odd, and vice versa – for $\sigma$ odd, $F(\sigma)$ is even. Finally $F \circ F=\mathrm{i} d_{S_{n}}$, i.e., $F$ is its own inverse. Thus if we restrict $F$ only on the subset ${$ even permutations , we will get a bijection between even and odd permutations.

发表回复

您的电子邮箱地址不会被公开。