Math4302 Modern Algebra (Lecture 10)

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 $n$.

$D_n\leq S_n$, 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 $\operatorname{ker}(\phi)=\{e\}$.

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 $g_1=g_2$.

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.

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