# Math4302 Modern Algebra (Lecture 10)

## Groups

### Group homomorphism

Recall the kernel of a group homomorphism is the set

$$
\operatorname{ker}(\phi)=\{a\in G|\phi(a)=e'\}
$$

<details>
<summary>Example</summary>

Let $\phi:(\mathbb{Z},+)\to (\mathbb{Z}_n,+)$ where $\phi(k)=k\mod n$.

The kernel of $\phi$ is the set of all multiples of $n$.

</details>

#### Theorem for one-to-one group homomorphism

$\phi:G\to G'$ is one-to-one if and only if $\operatorname{ker}(\phi)=\{e\}$

If $\phi$ is one-to-one, then $\phi(G)\leq G'$, $G$ is isomorphic ot $\phi(G)$ (onto automatically).

If $A$ is a set, then a permutation of $A$ is a bijection $f:A\to A$.

#### Cayley's Theorem

Every group $G$ is isomorphic to a subgroup of $S_A$ for some $A$ (and if $G$ is finite then $A$ can be taken to be finite.)

<details>
<summary>Example</summary>

$D_n\leq S_n$, so $A=\{1,2,\cdots,n\}$

---

$\mathbb{Z}_n\leq S_n$, (use the set of rotations) so $A=\{1,2,\cdots,n\}$ $\phi(i)=\rho^i$ where $i\in \mathbb{Z}_n$ and $\rho\in D_n$

---

$GL(2,\mathbb{R})$. Set $A=\mathbb{R}^2$, for every $A\in GL(2,\mathbb{R})$, let $\phi(A)$ be the permutation of $\mathbb{R}^2$ induced by $A$, so $\phi(A)=f_A:\mathbb{R}^2\to \mathbb{R}^2$, $f_A(\begin{pmatrix}x\\y\end{pmatrix})=A\begin{pmatrix}x\\y\end{pmatrix}$

We want to show that this is a group homomorphism.

- $\phi(AB)=\phi(A)\phi(B)$ (it is a homomorphism)

$$
\begin{aligned}
    f_{AB}(\begin{pmatrix}x\\y\end{pmatrix})&=AB\begin{pmatrix}x\\y\end{pmatrix}\\
    &=f_A(B\begin{pmatrix}x\\y\end{pmatrix})\\
    &=f_A(f_B(\begin{pmatrix}x\\y\end{pmatrix}))\\
    &=(f_A\circ f_B)(\begin{pmatrix}x\\y\end{pmatrix})\\
\end{aligned}
$$

- Then we need to show that $\phi$ is one-to-one.

It is sufficient to show that $\operatorname{ker}(\phi)=\{e\}$.

Solve $f_A(\begin{pmatrix}x\\y\end{pmatrix})=\begin{pmatrix}x\\y\end{pmatrix}$, the only choice for $A$ is the identity matrix.

Therefore $\operatorname{ker}(\phi)=\{e\}$.

</details>

<details>
<summary>Proof for Cayley's Theorem</summary>

Let $A=G$, for every $g\in G$, define $\lambda_g:G\to G$ by $\lambda_g(x)=gx$.

Then $\lambda_g$ is a **permutation** of $G$. (not homomorphism)

- $\lambda_g$ is one-to-one by cancellation on the left.
- $\lambda_g$ is onto since $\lambda_g(g^{-1}y)=y$ for every $y\in G$.

We claim $\phi: G\to S_G$ define by $\phi(g)=\lambda_g$ is a group homomorphism that is one-to-one.

First we show that $\phi$ is homomorphism.

$\forall x\in G$

$$
\begin{aligned}
    \phi(g_1)\phi(g_2)&=\lambda_{g_1}(\lambda_{g_2}(x))\\
    &=\lambda_{g_1g_2}(x)\\
    &=\phi(g_1g_2)x\\
\end{aligned}
$$

This is one to one since if $\phi(g_1)=\phi(g_2)$, then $\lambda_{g_1}=\lambda_{g_2}\forall x$, therefore $g_1=g_2$.

</details>

### Odd and even permutations

#### Definition of transposition

A $\sigma\in S_n$ is a transposition is a two cycle $\sigma=(i j)$

Fact: Every permutation in $S_n$ can be written as a product of transpositions. (may not be disjoint transpositions)

<details>
<summary>Example of a product of transpositions</summary>

Consider $(1234)=(14)(13)(12)$.

In general, $(i_1,i_2,\cdots,i_m)=(i_1i_m)(i_2i_{m-1})(i_3i_{m-2})\cdots(i_1i_2)$

This is not the unique way.

$$
(12)(34)=(42)(34)(23)(12)
$$

</details>

But the parity of the number of transpositions is unique.

#### Theorem for parity of transpositions

If $\sigma\in S_n$ is written as a product of transposition, then the number of transpositions is either always odd or even.

#### Definition of odd and even permutations

$\sigma$ is an even permutation if the number of transpositions is even.

$\sigma$ is an odd permutation if the number of transpositions is odd.