# Random math: 13

Source of most content: Gunhee Cho

# Propostions

$\text{Given a group } G,$

1. $\emptyset \ne H, K \leqslant G$, then $HK= \{ hk \mid h \in H, k \in K \} \leqslant G$ iff $HK=KH$.
2. If $H \trianglelefteq G$ and $K \leqslant G$, then $HK=KH$.
3. If $H \trianglelefteq G$ and $K \trianglelefteq G$, then $HK \trianglelefteq G$.
4. If $H \trianglelefteq G$, $K \trianglelefteq G$ and $H\cap K = \{e \}$, then $HK \cong H \times K = \{(h,k) \mid h \in H, k\in K \}$, where $HK$ is internal direct product and $H\times K$ is external direct product.

### Proof

• $(\Rightarrow)$ Let $HK \leqslant G$. Take any $hk \in HK$. Since $HK \leqslant G$, $(hk)^{-1} = k^{-1}h^{-1} \ in HK$, which implies $k^{-1}h^{-1} = h_0 k_0$ for some $h_0 \in H, k_0 \in K$. Therefore, $(k^{-1}h^{-1})^{-1} = hk = (h_0 k_0)^{-1} = k_0^{-1} h_0^{-1} \in KH$, which implies $HK \subseteq KH$. Similarly, we can show that $KH \subseteq HK \leadsto KH=HK.$
• $(\Leftarrow)$ Suppose $HK=KH$. Let $h_0k_0, h_1k_1 \in HK$. Then, $h_0 k_0 (h_1 k_1)^{-1} = h_0 k_0 k_1^{-1} h_1^{-1}.$ Since $HK=KH,\ k_0 k_1^{-1} h_1^{-1} = h_2 k_2$ for some $h_2 \in H, k_2 \in K \leadsto h_0 k_0 k_1^{-1} h_1^{-1} = h_0 h_2 k_2 \in HK.$ Therefore $HK \leqslant G$, since $h_0 k_0 (h_1 k_1)^{-1}\in HK$.
1. Let $H \trianglelefteq G$ and $K \leqslant G$. Take any $h_0 k_0 \in HK$. By assumption, $h_0 k_0 = k_0 k_0^{-1} h_0 k_0 \in KH$. Also, for any $k_1 h_1 \in KH, k_1h_1 = k_1h_1k_1^{-1} k_1 \in HK$. Therefore, $HK=KH$.

2. Let $H \trianglelefteq G, K \trianglelefteq G$. Then, $HK \leqslant G$ by (2) and (1). Take any $h_0 k_0 \in HK$ and $g \in G$. Then, $g_0 h_0 k_0 g^{-1} = gh_0 g^{-1} g k_0 g^{-1} \in HK$, which implies that $HK \trianglelefteq G$.

3. Let $H, K \trianglelefteq G$ and $H\cap K = \{e \}$. Then, by (3) $HK \trianglelefteq G$. Define $f: H\times K \rightarrow HK \ ((h,k)\mapsto hk)$, which is well-defined. To show $HK \cong H\times K$:
• (group homomorphism) Take any $(h_0, k_0), (h_1, k_1) \in H \times K.$ Then we need to show $f((h_0,k_0)(h_1,k_1))=f(h_0,k_0)f(h_1,k_1) \Leftrightarrow f(h_0h_1, k_0k_1)=h_0k_0h_1k_1 \Leftrightarrow h_1k_0h_1^{-1}k_0^{-1}=e$. Since $K \trianglelefteq G, (h_1 k_0 h_1^{-1})k_0^{-1}\in K$ and $H\trianglelefteq G \leadsto h_1 (k_0 h_1^{-1}k_0^{-1}) \in H$, therefore $h_1k_0h_1^{-1}k_0^{-1} \in K \cap H = \{ e\} \leadsto$ $f$ is a group homomorphism.
• (injective) $f(h_1, k_1) = f(h_2, k_2) \Leftrightarrow h_1k_1 = h_2 k_2 \Leftrightarrow k_1k_2^{-1} (\in K) =h_1^{-1}h_2 (\in H) \in H\cap K = \{ e \}.$ Therefore, $h_1=h_2, k_1=k_2$.
• (surjective) Take any $hk \in HK$. Then, $f(h, k)=hk$.

# Direct product/sum

Def. Given an index set $I, \forall i \in I, \ G_i$ is a group. Define $\Pi_{i\in I} G_i := \{f:I \rightarrow \cup_{i\in I}G_i \mid f(i)\in G_i, \forall i \in I \}$ which is called direct product of $\{ G_i \}_{i\in I}$.

e.g.) $A \times B = \{ (a,b) \mid a\in A, b \in B \} = \{f: \{1, 2\} \rightarrow A\cup B \mid f(1)\in A, f(2) \in B \}$.

NB1; In a vector notation, $\Pi _{i\in I} G_i := \{ (a_i) _{i=I} \}$. Then we give a binary operation on $\Pi _{i\in I} G_i$: given $f,g \in \Pi _{i\in I} G_i$, $f * g: I \rightarrow \cup _{i\in I} G_i$. Then, $(f * g)(i) = f(i) * g(i),$ i.e., $\Pi _{i\in I} G _i$ is a group.

Def. Given an index set $I, G_i$ is a group for each $i\in I$. Define $\oplus _{i \in I} G _i := \{f:I \rightarrow \cup _{i\in I} G_i \mid f(i)\in G_i, \forall i, f(i)= e _i \text{ all but finitely many } i’s \}$, which is called direct sum of $\{ G _i \} _{i\in I}$ (element is $(a _1,a _2, \cdots, a_n, e, e, \cdots, e, \cdots)$). In particular, $\oplus _{i \in I} G_i \leqslant \Pi _{i\in I} G_i$.

NB2; If $\mid I \mid \leq \infty$, $\oplus_{i \in I}G_i = \Pi _{i\in I} G_i$.

# Symmetric group

Def. For $n \in \mathbb{N}, \ S_n := \{f: \{ 1, 2, \cdots, n \} \rightarrow \{ 1, 2, \cdots, n \} \mid f \text{ is bijective}\}$. Then, $(S_n, \circ)$ is called a symmetric group, where $\circ$ is composition of functions. And we call the element $f \in S_n$ a permutation.

NB3; $\mid S_n \mid = n!$.
NB4; Every permutation in $S_n$ can be written by disjoint cycles, such as $\sigma=\tau_1 \circ \cdots \circ \tau_r \ (r \geq 1)$.
NB5; In general, $(i_1 \ i_2 \ \cdots \ i_n) = (i_1 \ i_n) \circ \cdots \circ (i_1 \ i_2),$ where $(a \ b \ c)$ is cyclic notation; $a$ to $b$, $b$ to $c$ and $c$ to $a$.

Def. $(i, j)$ is called transposition.

NB6; Every cycle is composition of transpotitions, such as $(1 \ 3) = (1 \ 3)(2\ 4)(4\ 2)= (1 \ 3)(2\ 4)(4\ 2)(1\ 2)(2\ 1)$.

Def. Given $(S_n, \circ)$ a symmetric group, then $\sigma \in S_n$ is even (odd) if $\sigma$ is written by even (odd) number of transpositions.

Def. Define $sgn: (S_n, \circ) \rightarrow (\{ +1, -1 \}, \times) \ (\sigma \mapsto 1 \text{ if } \sigma \text{ is even, } 0 \text{ o.w.})$. Then, $A_n :=\ker (sgn) = \{ \sigma \in S_n \mid sgn(\sigma)=1 \} \trianglelefteq S_n$. We call $A_n$ an alternating subgroup of $S_n$.

NB7; $sgn$ is well-defined $(\because \sigma_1 = \sigma_2 \leadsto sgn(\sigma_1)= sgn(\sigma_2))$ and group homomorphism $(\because \forall \sigma_1, \sigma_2 \in S_n, sgn(\sigma_1 \circ \sigma_2)= sgn(\sigma_1)\times sgn(\sigma_2))$.

Def. $A=(a_{ij}) \in Mat_{n\times n}(F), \det A = \sum_{\sigma\in S_n} sgn(\sigma) \Pi_{i=1}^n a_{i\sigma(i)}$ is called determinant.

## Proposition

$\text{Let } \tau_1 \circ \tau_2 \cdots \circ \tau_k = Id, \text{ where } \tau_i \text{ is a transposition. Then, } k \text{ is even.} \nonumber$

### Proof

Take any integer $m \in \{ 1, \cdots, n\}.$ Let $j_m$ be the largest $k$ s.t. $m \in \tau_k$. For example, $\sigma=(1\ 2)(3\ 4)(2\ 3)(5\ 2) \leadsto j_1 =1$. Then, there are 4 possible cases: for distinct $x,y,m,n$:

1. $(m\ x)\circ (m\ x)=Id$.
2. $(m\ y)\circ (m\ x)=(m\ x)\circ (y\ x)$.
3. $(n\ x)\circ (m\ x)=(m\ n)\circ (x\ n)$.
4. $(n\ y)\circ (m\ x)=(m\ x)\circ (n\ y)$.

This implies that we can move $m$ to $j_m -1$, i.e., $m$ can be contained only in $j_m -1$th transposition. Therefore we can move $m$ until $j_m =0$ or $1$. Here, if $j_m =1$, this implies there is a transposition which is a contradiction that $\sigma=Id$. Since this holds for any $m$, this implies that $k=0$.

$\text{Let } \sigma\in S_n. \text{ Write } \sigma=\tau_1 \circ \tau_2 \circ \cdots \circ \tau_k = \tau'_1 \circ \tau'_2 \circ \cdots \circ \tau'_k, \text{ where } \tau_i, \tau'_i \text{ are transpositions. Then, } k+k' \text{ is even.} \nonumber$

$\implies$ Pairity conserves.

### Proof

Let $\sigma=\tau_1 \circ \tau_2 \circ \cdots \circ \tau_k = \tau’_1 \circ \tau’_2 \circ \cdots \circ \tau’_k$. Then, $(\tau_1 \circ \tau_2 \circ \cdots \circ \tau_k)^{-1} \circ \tau’_1 \circ \tau’_2 \circ \cdots \circ \tau’_k = Id$. Therefore, $k+k’$ is even.