# Math4302 Modern Algebra (Lecture 18) ## Groups ### Factor group Suppose $G$ is a group, and $H\trianglelefteq G$, then $G/H$ is a group. Recall from last lecture, if $\phi:G\to G'$ is a homomorphism, then $G/\ker(\phi)\simeq \phi(G)\leq G'$.
Example (continue from last lecture) $\mathbb{Z}\times\mathbb{Z}/\langle (1,1)\rangle\simeq \mathbb{Z}$ Take $\phi(a,b)=a-b$, this is a surjective homomorphism from $\mathbb{Z}\times\mathbb{Z}/\langle (1,1)\rangle$ to $\mathbb{Z}$ --- $\mathbb{Z}\times\mathbb{Z}/\langle (2,1)\rangle\simeq \mathbb{Z}$ where $\langle (2,1)\rangle=\{(2b,b)|b\in \mathbb{Z}\}$ Take $\phi(a,b)=a-2b$, this is a surjective homomorphism from $\mathbb{Z}\times\mathbb{Z}/\langle (2,1)\rangle$ to $\mathbb{Z}$ --- $\mathbb{Z}\times\mathbb{Z}/\langle (2,2)\rangle$ This should also be a finitely generated abelian group. ($\mathbb{Z}_2\times \mathbb{Z}$ actually) Take $\phi(a,b)=(a\mod 2,a-b)$ --- More generally, for $\mathbb{Z}\times \mathbb{Z}/\langle (a,b)\rangle$. This should be $\mathbb{Z}\times \mathbb{Z}_{\operatorname{gcd}(a,b)}$ Try to do section by gcd.
> - If $G$ is abelian, $N\leq G$, then $G/N$ is abelian. > - If $G$ is finitely generated and $N\trianglelefteq G$, then $G/N$ is finitely generated. #### Definition of simple group $G$ is simple if $G$ has no proper ($H\neq G,\{e\}$), normal subgroup. > [!TIP] > > In general $S_n$ is not simple, consider the normal subgroup $A_n$.
Example of some natural normal subgroups If $\phi:G\to G'$ is a homomorphism, then $\ker(\phi)\trianglelefteq G$. --- The **center** of $G$: $Z(G)=\{a\in G|ag=ga\text{ for all }g\in G\}$ $Z(G)\trianglelefteq G$. - $e\in Z(G)$. - $a,b\in Z(G)\implies abg=gab\implies ab\in Z(G)$. - $a\in Z(G)\implies ag=ga\implies a^{-1}\in Z(G)$. - If $g\in G, h\in Z(G)$, then $ghg^{-1}\in Z(G)$ since $ghg^{-1}=gg^{-1}h=h$. $Z(S_3)=\{e\}$, all the transpositions are not commutative, so $Z(S_3)=\{e\}$. $Z(GL_n(\mathbb{R}))$? continue on friday.