DiscMat assignment 8

exercise

Exercise 8.4

1)

Reflexive: $\forall a\in G(a\cdot e=a)$ and $e\in H$ (because $H$ is a subgroup of $G$, it must contains $e$) hence $\forall a\in G\exists h\in H(a\cdot h=a)$, (take $h=e$) so $\forall a\in G(a\sim a)$ $⟹\sim$ is reflexive

Symmetric: For any $a,b\in G$ that satisfies $a\sim b$ $\exists h\in H(a\cdot h=b)$ (definition of $\sim$) $⟹a\cdot h\cdot \hat{h}=b\cdot \hat{h}$ (the inverse of $h$ exists, because $H$ is a group and $h\in H$) $⟹a\cdot e=b\cdot \hat{h}$ $⟹b\cdot \hat{h}=a$ hence $\exists h\in H(b\cdot h=a)$, (take $h=\hat{h}$) ($\hat{h}\in H$, because $H$ is a group) Therefore, according to the definition of $\sim$, $b\sim a$ $⟹\forall a,b\in G(a\sim b\to b\sim a)$ $⟹\sim$ is symmetric

Transitive For any $a,b,c\in G$ that satisfies $a\sim b$ and $b\sim c$ $\exists h_1\in H(a\cdot h_1=b)$ and $\exists h_2\in H(b\cdot h_2=c)$ (definition of $\sim$) Let $h_3=h_1\cdot h_2$ and $h_3\in H$ (Since group $H$ is closed and $h_1\in H\wedge h_2\in H$, $h_3$ must also be in $H$) $⟹c=b\cdot h_2=a\cdot h_1\cdot h_2=a\cdot h_3$ $⟹\exists h\in H(a\cdot h=c)$ (take $h=h_3$) $⟹\forall a,b,c\in G(a\sim b\wedge b\sim c\to a\sim c)$ $⟹\sim$ is transitive

Summary Since $\sim$ is reflexive, symmetric and transitive, it is an equivalence relation.

2)

We prove $\pi (a\cdot b)=\pi (a^{\prime }\cdot b^{\prime })$ by showing that $a\cdot b\sim a^{\prime }\cdot b^{\prime }$ (If $a\cdot b\sim a^{\prime }\cdot b^{\prime }$, then they must be in the same equivalence class)

Since $a\sim a^{\prime }$ and $b\sim b^{\prime }$, there exist $h_1$ and $h_2$ such that $a\cdot h_1=a^{\prime }$ and $b\cdot h_2=b^{\prime }$ Therefore, $a^{\prime }\cdot b^{\prime }=a\cdot h_1\cdot b\cdot h_2$ $⟹a\cdot b\cdot h_1\cdot h_2=a^{\prime }\cdot b^{\prime }$ (since $G$ is an abelian group) Let $h_3=h_1\cdot h_2$ and $h_3\in H$ (Since group $H$ is closed and $h_1\in H\wedge h_2\in H$, $h_3$ must also be in $H$) $⟹\exists h\in H((a\cdot b)\cdot h=a^{\prime }\cdot b^{\prime })$ (take $h=h_3$) $⟹a\cdot b\sim a^{\prime }\cdot b^{\prime }$ $⟹\pi (a\cdot b)=\pi (a^{\prime }\cdot b^{\prime })$

3)

Associativity Let $[a{]}_{\sim },[b{]}_{\sim },[c{]}_{\sim }\in G/H$ $([a{]}_{\sim }\star [b{]}_{\sim })\star [c{]}_{\sim }\stackrel{def}{=}(\pi (a)\star \pi (b))\star \pi (c)=\pi (a\cdot b)\star \pi (c)=\pi ((a\cdot b)\cdot c)$ $=\pi (a\cdot (b\cdot c))=\pi (a)\star \pi (b\cdot c)=\pi (a)\star (\pi (b)\star \pi (c))=[a{]}_{\sim }\star ([b{]}_{\sim }\star [c{]}_{\sim })$

Neutral Element The neutral element $e^{\prime }$ in $⟨G/H;\star ⟩$ is $[e{]}_{\sim }$ (or denoted as $\pi (e)$) Let $\pi (a)\in G/H$ $\pi (a)\star \pi (e)\stackrel{def}{=}\pi (a\cdot e)=\pi (a)$ $\pi (e)\star \pi (a)\stackrel{def}{=}\pi (e\cdot a)=\pi (a)$ This means $\pi (a)\star \pi (e)=\pi (e)\star \pi (a)=\pi (a)$, which proves that $e^{\prime }$ is the neutral element in $⟨G/H;\star ⟩$

Inverse Let $a\in G$ and $\pi (a)\in G/H$ We show that the inverse $\widehat{\pi (a)}=\pi (\hat{a})$ ($\hat{a}$ exists because $G$ is a group, which means $\pi (\hat{a})$ exists) $\pi (a)\star \pi (\hat{a})=\pi (a\cdot \hat{a})=\pi (e)=e^{\prime }$ Well-definedness: To prove the inverse operation is well defined, we show: For $a,b\in G$ and $\pi (a),\pi (b)\in G/H$, $\pi (a)=\pi (b)\to \widehat{\pi (a)}=\widehat{\pi (b)}$ $\pi (a)=\pi (b)⟹a\sim b⟹\exists h\in H(a\cdot h=b)⟹\hat{a}\cdot \hat{h}=\hat{b}$ $⟹\hat{a}\sim \hat{b}⟹\pi (\hat{a})=\pi (\hat{b})⟹\widehat{\pi (a)}=\widehat{\pi (b)}$ Therefore the inverse operation is well defined.

4)

Yes, $\pi$ is a homomorphism

8.7 Kernel of an homomorphism

We first show that $ker(\varphi )=\{e_G\}⟹\varphi$ is injective: $\varphi (e_G)=e_H$ Let $a,b\in G$ and satisfy $\varphi (a)=\varphi (b)$ $⟹\varphi (a)\odot \widetilde{\varphi (b)}=e_H$ $⟹\varphi (a)\odot \varphi (\hat{b})=e_H=\varphi (e_G)$ (because $\widetilde{\varphi (b)}=\varphi (\hat{b})$, see side proof) $⟹\varphi (a+\hat{b})=e_H=\varphi (e_G)$ (property of homomorphism) Since $e_G$ is the only element in $G$ that satisfy $g\in G\wedge \varphi (g)=e_H$, we must have $a+\hat{b}=e_G$ $⟹a+\hat{b}+b=e_G+b$ $⟹a+e_G=b$ $⟹a=b$ $⟹\forall a,b\in G(\varphi (a)=\varphi (b)\to a=b)$ $⟹\varphi$ is injective side proof: $\varphi (b)\odot \varphi (\hat{b})=\varphi (b+\hat{b})=\varphi (e_G)=e_H$ $\varphi (\hat{b})\odot \varphi (b)=\varphi (\hat{b}+b)=\varphi (e_G)=e_H$ $⟹\varphi (\hat{b})$ is an inverse of $\varphi (b)$ $⟹\widetilde{\varphi (b)}=\varphi (\hat{b})$

We then show that $\varphi$ is injective $⟹ker(\varphi )=\{e_G\}$: $\varphi$ is injective $⟹$ for all $a,b\in G$ $\varphi (a)=\varphi (b)\to a=b$ Let’s assume some $g\in ker(\varphi )$ satisfies $g\in G,\varphi (g)=e_H$ Let $x\in G$ $e_H\odot \varphi (x)=\varphi (g)\odot \varphi (x)=\varphi (g+x)=\varphi (x)$ (property of homomorphism) $⟹g+x=x$ ($\varphi$ is injective) $\varphi (x)\odot e_H=\varphi (x)\odot \varphi (g)=\varphi (x+g)=\varphi (x)$ (property of homomorphism) $⟹x+g=x$ ($\varphi$ is injective) Therefore $x+g=g+x=x$ $g$ is a neutral element in $G$ $⟹g=e_G⟹ker(\varphi )=\{e_G\}$

Conclusion Since we proved the statement in both directions, the statement holds.

8.8 Conjugacy

We prove $T$ is a subgroup of $G$ by showing $\forall a,b\in T(a\star b\in H)$ and $\forall a\in T(\hat{a}\in T)$ We first show $\forall a,b\in T(a\star b\in T)$ Take $a,b\in T$ and $h\in H$ $⟹a\star h\star \hat{a}\in H\wedge b\star h\star \hat{b}\in H$

$⟹(a\star b)\star h\star \widehat{(a\star b)}\in H$

We then show that $\forall a\in T(\hat{a}\in T)$ Take $a\in T$ and $h\in H$ $⟹a\star h\star \hat{a}\in H$