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# Math4302 Modern Algebra (Lecture 9)
## 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.