# Random math: 09

Source of most content: Gunhee Cho

# Subgroup

Def. Given a group $(G, *), H \subseteq G$ is called a subgroup of $G$ if $(H, *)$ is a group and denoted by $H \leqslant G$.

e.g.) $(\mathbb{Q}, +) \leqslant (\mathbb{R}, +)$,

NB1; Let $H \leqslant (G, *)$. Since group is a monoid, $\exists! e_G \in (G, *), \exists! e_H \in (H, *)$ where $e$ implies identity. Then, by uniqueness of identity, $e_H = e_G$.
NB2; Let $H \subseteq G$. Then, $H \leqslant G$ iff:

1. $\forall a,b \in H, \ a * b \in H$.
2. $\forall c \in H, \ c^{-1} \in H$.

### Proof

1. $(\Rightarrow)$ Since $(H, *)$ is also a group, $\forall a,b \in H, a * b \in H$ and $\forall c \in H, \exists d \in H$ s.t. $c * d = d * c = e \in H$, where $d = c^{-1}$.
2. $(\Leftarrow)$ By (1) and (2), $\exists e \in H$, which implies $(H, *)$ is a monoid. Therfore, $(H, *)$ is also a group by (2).

Def. Given a group $(G, *), S \subseteq G$. Define $span(S):= < S > := \cap_{S\subseteq H \leqslant G} H$ which is called a subgroup generated by $S$.

e.g.) Let $(G, +)$ be an abelian group. Pick $x\in G \leadsto \{x\} \subseteq G$, then $<\{x\}> = \{ nx \mid n \in \mathbb{Z} \}$.

NB3; $< S>$ is equivalent to the smallest subgroup containing $S$.
NB4; Given a group $(G, *), \forall i \in I, H_i \leqslant G$. Then, $\cap_{i\in H} H_i \leqslant G$.

### Proof

Let $H_1, H_2 \leqslant (G, *)$. Then, $\forall a,b \in H_1 \cap H_2 := H$, $a,b \in H_1 \land a,b \in H_2 \leadsto a, b^{-1} \in H_1 \land a, b^{-1} \in H_2$ $\Leftrightarrow a * b^{-1} \in H$.

# Homomorphism

Def. Given two groups $(G, *), (H, \circ)$, we call $f:G\rightarrow H$ is a group homomorphism if $\forall g_1, g_2\in G, \ f(g_1 * g_2) = f(g_1) \circ f(g_2)$. Given a group-homomorphism $f:G \rightarrow H$:

1. $f$ is called a monomorphism if $f$ is injective.
2. $f$ is called a epimorphism if $f$ is surjective.
3. $f$ is called a isomorphism if $f$ is bijective. In this case, it is denoted by $G \stackrel{f}{\cong} H$ and this implies that $G$ and $H$ has same group structure.

Def. Given two rings $(R_1, +_1, \times_1), (R_2, +_2, \times_2)$, $f:R_1 \rightarrow R_2$ is a ring homomophism if $\forall r_1, r_2 \in R_1, f(r_1 +_1 r_2) = f(r_1) +_2 f(r_2)$ and $f(r_1 \times_1 r_2)= f(r_1) \times_2 f(r_2)$.

Def. Given a ring $R$, and $R$-modules $(M_1, +_1, \cdot_1), (M_2, +_2, \cdot_2)$, $f:M_1 \rightarrow M_2$ is a module homomorphism if $\forall m_1, m_2 \in M_1, f(m_1 +_1 m_2) = f(m_1) +_2 f(m_2)$ and $\forall r \in R, \forall m \in M,$ $f(r \cdot_1 m) = r \cdot_2 f(m)$.

From this definition, we can notice that this is a generalization of linear transformation between VS.

Def. Given two vector spaces $V,W/F$, we call $\Phi:V\rightarrow W$ is a linear transformation if $\Phi$ is module homomorphism over field $F$. (note that field is a ring)

## Propostion

$\text{Given }f:(G, *) \rightarrow (H, \circ) \text{ a group homomorphism. Then, following holds:} \nonumber$
1. $f(e_G)=e_H$
2. $\forall g \in G, f(g^{-1})=(f(g))^{-1}$.
3. $f^{-1}(\{ e_H \})=\{ g \in G \mid f(g) = e_H \} := \ker f \leqslant G$, the kernel of $f$.
4. $\ker f = \{ e_G\}$ iff $f$ is injective.
5. $\text{Im} f = \{f(g) \mid g \in G \} \leqslant H$.

### Proof

1. $f(e_G * e_G) = f(e_G) \circ f(e_G)$. Since $f(e_G) \in H, \ \exists f^{-1}(e_G)\in H \leadsto f(e_G)\circ f^{-1}(e_G) = e_H = f(e_G).$
2. $\forall g \in G, \ f(g * g^{-1})= f(e_G) = f(g) \circ f(g^{-1}) = e_H \leadsto f(g^{-1})=(f(g))^{-1}$.
3. $\forall g_1, g_2 \in \ker f, f(g_1 * g_2) = f(g_1) \circ f(g_2) = e_H \circ e_H = e_H \leadsto g_1 * g_2 \in \ker f$. And $\forall g \in \ker f, f(g) = e_H \leadsto f(g)\circ f(g) = e_H \leadsto \exists f^{-1}(g)=f(g^{-1}) = e_H \leadsto g^{-1} \in \ker f$. Therefore, $\ker f \leqslant G$.
• $(\Rightarrow)$ Let $g_1, g_2 \in G$ s.t. $f(g_1) = f(g_2)$ $\leadsto f(g_1 * g_2^{-1})=f(g_1)\circ f(g_2^{-1}) =f(g_1)\circ f^{-1}(g_2) = f(g_1)\circ f^{-1}(g_1)=e_H$. This implies that $g_1 g_2^{-1} \in \ker f$. Since $\ker f = \{ e_G \}$, $g_1 g_2^{-1} = e_G \leadsto g_1 = g_2$, i.e., $f$ is injective.
• $(\Leftarrow)$ By (1), $e_G \in \ker f$. Since $f$ is injective, if any $g$ satisfies $f(g)=e_H$, then $g=e_G \leadsto \ker f = \{ e_G \}$.
4. $\forall g_1, g_2 \in G, f(g_1) \circ f(g_2) = f(g_1 * g_2) \in \text{Im} f, \ \because g_1 * g_2 \in G$. And $\forall g \in G, f^{-1}(g) \in \text{Im} f$, since $f^{-1}(g) = f(g^{-1})$ and $g^{-1} \in G$. Therefore, $\text{Im}f \leqslant H$.

NB5; Let $\Phi: V \rightarrow W$ be a linear transformation over $F$. Then, $\ker \Phi = \{ v \in V \mid \Phi(v) =0 \} \leqslant V, \text{Im}\Phi \leqslant W$. We call $dim \ker \Phi$ nullity of $\Phi$ and $dim \text{Im}\Phi$ rank of $\Phi$ and the sum of two is equivalent to $dimV$ by rank-nullity theorem.

# Cyclic group

Def. For any group $G$, take any $x \in G$. Then, $span(\{x\}) := \cap_{x\in H \leqslant G} H = \{ x^n \mid n \in \mathbb{Z} \}$ is called cyclic group generated by $x$ and denoted by $< x >$.

e.g.) $(\mathbb{Z}, +) = <1>=<-1>$.
e.g.) $(\mathbb{Z}/n\mathbb{Z}, +)=<[0],[1],\cdots,[n-1]>=<[1]>,$ where $[ k ]=\{k+nm \mid m \in \mathbb{Z}\} := k + n\mathbb{Z}$.

## Proposition

$\text{Given a group } (G,*), \text{ for any } x \in G, \text{<}x\text{>} \text{ is either isomorphic to } (\mathbb{Z}, +) \text{ or isomorphic to } (\mathbb{Z}/n\mathbb{Z}, +) \text{ for some }n. \nonumber$

### Proof

There are two cases (1) $\nexists i\ne j, i,j\in \mathbb{Z}$ s.t. $x^i = x^j$ and (2) $\exists i\ne j, i,j \in \mathbb{Z}$ s.t. $x^i = x^j$.

1. Define $\Phi: < x > \rightarrow \mathbb{Z} \ (x^n \mapsto n)$. By Asm. (1), $\Phi$ is well-defined, i.e., if $x^n=x^m \Leftrightarrow n=m$.
• $\Phi(x^n * x^m)=\Phi(x^{n+m}) = n+m = \Phi(x^n) + \Phi(x^m)$, which implies that $\Phi$ is a group homomorphism.
• If $x^n \ne x^m$, then, $n \ne m$, i.e., $\Phi$ is injective.
• Take any $n\in \mathbb{Z}$, then $\Phi(x^n)=n$, i.e., $\Phi$ is surjective.
$\implies < x> \stackrel{\Phi}{\cong} \mathbb{Z}$.
2. Let $i,j \in \mathbb{Z}$ s.t. $x^i = x^j$. Assume $i > j \leadsto x^i (x^j)^{-1} = e \Leftrightarrow x^{i-j}=e$. Let $S = \{ n \in \mathbb{N} \mid x^n = e \}$. Then, $S \ne \emptyset$. Therefore, by W.O. principle, $\exists n_0 = \min S$. Define $\varphi: < x > \rightarrow \mathbb{Z}/n_0\mathbb{Z} \ (x^n \mapsto [n])$. If $\varphi$ is well-defined, $x^n = x^m$ implies $\varphi(x^n)=[n]=\varphi(x^m)=[m]$, i.e., $n-m=n_0 k$ for some $k \in \mathbb{Z}$. Assume $n-m = n_0 k + l$, where $0 < l < n_0$. Then, $x^{n-m}=x^{n_0 k + l}= x^{n_0 k}x^l=x^l =e$, which is a contradiction since $n_0$ is a minimum that satisfies $x^n_0 = e$. Therefore, $\varphi$ is well-defined.
• $\forall n,m \in \mathbb{Z} \varphi(x^n * x^m)=\varphi(x^{n+m})=[n+m]=\varphi(x^n)+\varphi(x^m)$, which implies that $\varphi$ is a group homomorphism.
• If $\varphi(x^n)=\varphi(x^m)$, i.e., $[n]=[m] \leadsto \exists l_0 \in \mathbb{Z}$ s.t. $n-m=n_0 l_0$. This implies $x^n = x^{m+n_0 l_0}=x^m$, i.e., $\varphi$ is injective.
• Take any $[n] \in \mathbb{Z}/n_0\mathbb{Z}, \varphi(x^n)=[n]$, i.e., $\varphi$ is surjective.
$\implies < x> \stackrel{\varphi}{\cong} \mathbb{Z}/n\mathbb{Z}$.

$\text{Let } G = < x >, x \in G. \text{If } H \leqslant G, \text{ then } H \text{ is cyclic}. \text{ i.e., subgroup of cyclic group is cyclic.} \nonumber$

### Proof

Let $S = \{ n \in \mathbb{N} \mid x^n \in H \} \ne \emptyset$. By W.O., $\exists m_0 = \min S$. Then, we need to show $H=< x^{m_0} > = \{ (x^{m_0})^k \mid k \in \mathbb{Z}\}$.

1. $(\supseteq)$ $x^{m_0} \in H \leadsto \forall k \in \mathbb{Z} / (x^{m_0})^k \in H$ since $H$ is a group.
2. $(\subseteq)$ Let $x^n \in H,$ for some $n\in\mathbb{Z}$. Then, by division, $\exists q \in \mathbb{Z}, r\in \{0, \cdots, m_0 -1 \}$ s.t. $n=m_0 q + r$, i.e., $r = n- m_0 q$. Then, $x^r = x^{n-m_0 q} = x^n * ((x^{m_0})^k)^{-1} \in H$, i.e., $x^r \in H$ for $r \leq m_0 -1 \leq m_0$ which is a contradiction unless $r=0$. Therefore, $x^n \in < x^{m_0} >$.