ALG2_EndSem

Vasudeva S N

Question 4

Part a

Let $G=(\mathbb{R},+)$. Show that every subgroup of $G$ is either dense or discrete.

Let $H\le G$. Assume $H$ is not discrete, that is, there does not exist $ϵ>0$ such that all non-zero elements of $G$ have absolute value greater than $ϵ$. We have to show that $H$ is dense in $G$. Let $g\in G\setminus H$. We will show that $g$ is a limit point of $H$. Fix $ϵ>0$. There must exist $h\in H$ with $|h|\le ϵ$. Thus, there must exist a multiple $kh\in (g-ϵ,g+ϵ)$, where $k\in \mathbb{N}$. For the $ϵ$-sequence ${ϵ}_n=\frac{1}{n}$, let the corresponding sequence of $h$s be $(h_n)$. It follows that $(h_n)\to g$. Thus, $H$ is dense in $G$.

Part b

Let $H$ be the subgroup generated by $1$ and $\sqrt{2}$. By part (a), we only have to show that $H$ is not discrete. Note that $\sqrt{2}-1\in H$, $\sqrt{2}-1<1$, and $(\sqrt{2}-1{)}^n\in H$ for all $n\in \mathbb{N}$. Further, the $(\sqrt{2}-1{)}^n\to 0$ as $n\to 0$. Thus, for any $ϵ>0$, we can find $n\in \mathbb{N}$ such that $(\sqrt{2}-1{)}^n<ϵ$. Thus, $H$ is not discrete.

Part c

We have to show that if $H$ is not dense, it is cyclic. Suppose $H$ is discrete and non-trivial. If $a,b\in H$, $a-b$ is also in $H$, and hence $|a-b|>ϵ$. Thus, any two elements of $H$ are separated by at least $ϵ$, and thus any interval in $\mathbb{R}$ must contain finitely many elements of $H$. Let $a$ be any positive element in $H$. Choose the smallest element $b$ in $(0,a]$, which we can do since this interval must contain finitely many elements of $H$.

We will now show that $H=\mathbb{Z}b$. It is clear that $\mathbb{Z}b\subseteq H$. Let $c\in H$. $c=kb+r$, where $k\in \mathbb{Z}$, and $0\le r<b$. Then, $r=c-kb\in H$, and since cannot lie in $(0,b)$, it must be $0$. Thus, $c=kb$, and $H=\mathbb{Z}b$.

Question 5

Consider the case when $a=e$, that is, when $H$ is a Sylow-$p$ subgroup. Then,

$$\begin{array}{rl}\frac{|N_G(H)|-|G|}{|H|}& =\frac{|G|}{|H|}\left(\frac{1}{|{\text{Syl}}_p(G)|}-1\right)\\ & =m\left(\frac{1}{1+kp}-1\right)\\ & =p\left(-\frac{km}{1+kp}\right).\end{array}$$

Note that $\frac{km}{1+kp}$ must be an integer, since $1+kp$ does not divide $p$ and the quantity on the LHS is an integer. Thus, $p$ divides the LHS, and the claim is proved.

Next, let $a<e$. It is sufficient to show that $H$ is a proper subgroup of $N_G(H)$. By Sylow’s theorem, we can find a Sylow-$p$ subgroup $P$ such that $H<P$.

Claim 1(Claim).

If $P$ is any $p$-group and $H<P$, then $H<N_P(H)$.

Proof.

We will prove the claim by induction on $|P|$. If $|P|=1$, the claim is vacuously true. For $|P|>1$, we know that $Z(P)$ is non-trivial. If $Z(P)$ is not a subgroup of $H$, then any $z\in Z(P)\setminus H$ demonstrates that $N_G(H)$ contains elements not in $H$. If $Z(P)\le H$, Consider the groups $H^{\prime }=H/Z(P)$ and $P^{\prime }=P/Z(P)$. Since $H^{\prime }<P^{\prime }$, by the induction hypothesis, $H^{\prime }<N_{P^{\prime }}(H^{\prime })$. However, $N_{P^{\prime }}(H^{\prime })=N_P(H)/Z(P)$, so we have

$$\frac{H}{Z(P)}<\frac{N_P(H)}{Z(P)},$$

which implies $H<N_P(H)$.□

Thus, $H$ is a proper subgroup of $N_P(H)$, and since $N_P(H)\subseteq N_G(H)$, we have $H<N_G(H)$. Hence, $p$ divides the quantity $\frac{|N_G(H)|-|G|}{|H|}$, and we are done.

Question 7

Part b

If $H$ and $K$ are subgroups of $G$, we know that either $H◃G$ or $K◃G$ implies $HK\le G$. Thus, by Part (a), $HK<G$. Further, $|HK|=|H||K|/|H\cap K|$. Now, $|H\cap K|=1$, since $|H|$ and $|K|$ are both prime. Thus, $|HK|=pq$. This makes $HK$ a subgroup of index $2$ in $G$, and hence, $HK◃G$.

From our analysis of groups of order $pq$, if $p$ does not divide $q-1$, $HK$ is a cyclic subgroup. However, if $p$ does divide $q-1$, a non-cyclic group of order $pq$ can be constructed using the semidirect product, since a non-trivial homomorphism from $H$ to $\text{Aut}(K)$ is now possible ($|\text{Aut}(K)|=q-1$).

Part d

If $p$ does not divide $q-1$, then $HK={\mathbb{Z}}_{pq}$ is a cyclic subgroup of $G$. Also note that $G/HK\cong {\mathbb{Z}}_2$. Also, since $p$ and $q$ are prime,

$$\begin{array}{rl}\text{Aut}({\mathbb{Z}}_{pq})& =\text{Aut}({\mathbb{Z}}_p\times {\mathbb{Z}}_q)\\ & =\text{Aut}({\mathbb{Z}}_p)\times \text{Aut}({\mathbb{Z}}_q)\\ & ={\mathbb{Z}}_p^{\times }\times {\mathbb{Z}}_q^{\times }\end{array}$$

Consider homomorphisms $\varphi :{\mathbb{Z}}_2\to \text{Aut}({\mathbb{Z}}_{pq})={\mathbb{Z}}_p^{\times }\times {\mathbb{Z}}_q^{\times }$. Note that ${\mathbb{Z}}_p^{\times }\cong {\mathbb{Z}}_{p-1}$ and ${\mathbb{Z}}_q^{\times }\cong {\mathbb{Z}}_{q-1}$, and since $p-1$ and $q-1$ are even, both ${\mathbb{Z}}_{p-1}$ and ${\mathbb{Z}}_{q-1}$ have exactly one element of order $2$ each, namely $\frac{p-1}{2}$ and $\frac{q-1}{2}$. Their counterparts in ${\mathbb{Z}}_p^{\times }$ and ${\mathbb{Z}}_q^{\times }$ would be $-1$ and $-1$. So, if we let $z$ represent the nonidentity element in ${\mathbb{Z}}_2$, we have four possible homomorphisms:

$$\begin{array}{rl}{\phi }_1& :z\mapsto (1,1),\\ {\phi }_2& :z\mapsto (-1,1),\\ {\phi }_3& :z\mapsto (1,-1),\\ {\phi }_4& :z\mapsto (-1,-1).\end{array}$$

${\phi }_1$ corresponds to the direct product ${\mathbb{Z}}_{pq}\times {\mathbb{Z}}_2\cong {\mathbb{Z}}_{2pq}$.

Question 8

Part a

For $g,h\in G$, if $g(H\cap K)=h(H\cap K)$, then $h^{-1}g\in H\cap K$, which implies $h^{-1}g\in H$ and $h^{-1}g\in K$.

$$\varphi (g(H\cap K))=(gH,gK)=(hH,hK)=\varphi (h(H\cap K)).$$

Thus, $\varphi$ is well defined. Further,

$$\begin{array}{rl}\varphi (gh(H\cap K))& =(ghH,ghK),\\ & \\ \varphi (g(H\cap K))\varphi (h(H\cap K))& =(gH,gK)(hH,hK)\\ & =(ghH,ghK).\end{array}$$

Thus, $\varphi$ is an homomorphism.

Part b

$$\begin{array}{rl}\ker \varphi & =\left\{g(H\cap K)\in \frac{G}{H\cap K}:gH=H,gK=K\right\}\\ & =\left\{g(H\cap K)\in \frac{G}{H\cap K}:g\in H\cap K\right\}\\ & =\{(H\cap K)\}\\ & =\left\{1_{\frac{G}{H\cap K}}\right\}.\end{array}$$

Thus, $\varphi$ is injective.

Part c

Let $aH=a^{\prime }H$ and $bK=b^{\prime }K$. Then, $a^{\prime -1}a\in H$, and $b^{\prime -1}b\in K$. So, $a^{\prime -1}a{b}^{\prime }{b}^{-1}=b^{\prime }{a}^{\prime -1}a{b}^{-1}\in HK$, and hence $a{b}^{-1}HK=a^{\prime }{b}^{\prime -1}HK$. Thus, $\psi$ is well defined. Further for any $a,a^{\prime },b,b^{\prime }\in G$,

$$\begin{array}{rl}\psi (a{a}^{\prime }H,b{b}^{\prime }K)& =a{a}^{\prime }{b}^{\prime -1}{b}^{-1}HK,\\ & \\ \psi (aH,bK)\psi (a^{\prime }H,b^{\prime }K)& =(a{b}^{-1}HK)(a^{\prime }{b}^{\prime -1}HK)\\ & =a{b}^{-1}{a}^{\prime }{b}^{\prime -1}HK\\ & =a{a}^{\prime }{b}^{\prime -1}{b}^{-1}HK.\end{array}$$

Thus, $\psi$ is a homomorphism.

Part d

For any $gHK\in G/HK$, $\psi (gH,K)=gHK$.

Part e

Clearly, $\mathrm{Im}\varphi \subseteq \ker \psi$, since $\psi (\varphi (g(H\cap K)))=HK$. Next, assume $(aH,bK)\in \ker \psi$. Then, $a{b}^{-1}\in HK$, so $a{b}^{-1}=hk$ for some $h\in H$ and $k\in K$. Define $c=a{h}^{-1}=bk$.

$$(aH,bK)=(a(h^{-1}H),b(kK))=(cH,cK)\in \mathrm{Im}\varphi .$$

Thus, $\ker \psi \subseteq \mathrm{Im}\varphi$, and consequently $\ker \psi =\mathrm{Im}\varphi$.

Part f

Theorem 2(Chinese remainder theorem).

If positive integers $m$ and $n$ are co-prime,

$$\frac{\mathbb{Z}}{mn\mathbb{Z}}\cong \frac{\mathbb{Z}}{m\mathbb{Z}}\times \frac{\mathbb{Z}}{n\mathbb{Z}}.$$

Let $G=\mathbb{Z}$, $H=m\mathbb{Z}$, $K=n\mathbb{Z}$. Then, we have

$$\frac{\mathbb{Z}}{\text{lcm}(m,n)\mathbb{Z}}=\frac{\mathbb{Z}}{mn\mathbb{Z}}\stackrel{\varphi }{\to }\frac{\mathbb{Z}}{m\mathbb{Z}}\times \frac{\mathbb{Z}}{n\mathbb{Z}}\stackrel{\psi }{\to }\frac{\mathbb{Z}}{m\mathbb{Z}+n\mathbb{Z}}.$$

If $m$ and $n$ are co-prime, $m\mathbb{Z}+n\mathbb{Z}=\mathbb{Z}$, so $\frac{\mathbb{Z}}{m\mathbb{Z}+n\mathbb{Z}}$ is the trivial group. It follows that $\ker \psi =\mathbb{Z}/m\mathbb{Z}\times \mathbb{Z}/n\mathbb{Z}$. Since $\varphi$ is an injective homomorphism, $\mathbb{Z}/mn\mathbb{Z}\cong \mathrm{Im}\varphi$, and since $\mathrm{Im}\varphi =\ker \psi$, $\mathbb{Z}/mn\mathbb{Z}\cong \mathbb{Z}/m\mathbb{Z}\times \mathbb{Z}/n\mathbb{Z}$.

Conversely, if $\varphi$ is an isomorphism, $\ker \psi =\mathrm{Im}\varphi =\mathbb{Z}/m\mathbb{Z}\times \mathbb{Z}/n\mathbb{Z}$, so $\mathrm{Im}\psi =\left\{1_{\frac{\mathbb{Z}}{m\mathbb{Z}+n\mathbb{Z}}}\right\}$. Since $\psi$ is surjective, $\frac{\mathbb{Z}}{m\mathbb{Z}+n\mathbb{Z}}=\mathrm{Im}{\varphi }_2$. This implies $m$ and $n$ generate $\mathbb{Z}$, which implies $\text{gcd}(m,n)=1$ or $\text{lcm}(m,n)=mn$.