| title |
author |
header-includes |
abstract |
Abstract Algebra by Pinter, Chapter 15 |
Amir Taaki |
\usepackage{amsmath}
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Chapter 15 on Quotients |
Let $G = \mathbb{Z}_10, H = {0, 5}$. Explain why $G/H \cong \mathbb{Z}_5$
Elements of $G/H$:
\begin{align*}
H + 0 = {0, 5} \
H + 1 = {1, 6} \
H + 2 = {2, 7} \
H + 3 = {3, 8} \
H + 4 = {4, 9}
\end{align*}
$G/H \cong \mathbb{Z}_5$ because let the isomorphism $f(Hx) = x$ then $f(Hx \cdot Hy) = f(Hx)f(Hy)$.
Let $G = S_3$ and $H = {\epsilon, \beta, \delta}$
$\epsilon = \left(\begin{smallmatrix}
1 & 2 & 3\
1 & 2 & 3
\end{smallmatrix}\right)$
$\alpha = \left(\begin{smallmatrix}
1 & 2 & 3\
1 & 3 & 2
\end{smallmatrix}\right)$
$\beta = \left(\begin{smallmatrix}
1 & 2 & 3\
3 & 1 & 2
\end{smallmatrix}\right)$
$\gamma = \left(\begin{smallmatrix}
1 & 2 & 3\
2 & 1 & 3
\end{smallmatrix}\right)$
$\delta = \left(\begin{smallmatrix}
1 & 2 & 3\
2 & 3 & 1
\end{smallmatrix}\right)$
$\kappa = \left(\begin{smallmatrix}
1 & 2 & 3\
3 & 2 & 1
\end{smallmatrix}\right)$
Elements of the quotient group:
\begin{align*}
H = H\epsilon = {\epsilon, \beta, \delta} \
H\alpha = {\alpha, \kappa, \gamma} \
\end{align*}
Let $G = D_4$ and $H = {R_0, R_2}$
Elements of $G/H$:
\begin{align*}
H = {R_0, R_2} \
HR_1 = {R_1, R_3} \
HR_4 = {R_4, R_5} \
HR_6 = {R_6, R_7}
\end{align*}
| Symbol |
Transform |
| $R_0$ |
Identity |
| $R_1$ |
Rotate 90 |
| $R_2$ |
Rotate 180 |
| $R_3$ |
Rotate 270 |
| $R_4$ |
Flip left diagonal |
| $R_5$ |
Flip right diagonal |
| $R_6$ |
Flip horizontal |
| $R_7$ |
Flip vertical |
Let $G = D_4$ and $H = {R_0, R_2, R_4, R_5}$. Elements are $H, HR_1$.
Let $G = \mathbb{Z}_4 \times \mathbb{Z}_2, H = <(0, 1)>$.
\begin{align*}
H = {(0, 0), (0, 1)} \
H + (1, 0) = {(1, 0), (1, 1)} \
H + (2, 0) = {(2, 0), (2, 1)} \
H + (3, 0) = {(3, 0), (3, 1)} \
\end{align*}
Let $G = P_3, H = {\emptyset, {1}}$.
\begin{align*}
H = {\emptyset, {1}} \
H \cap {2} = {{2}, {1, 2}} \
H \cap {3} = {{3}, {1, 3}} \
H \cap {2, 3} = {{2, 3}, {1, 2, 3}} \
\end{align*}
$H = {(x, 0): x \in \mathbb{R}}$
For any $a \in H$ and $x \in G = \mathbb{R} \times \mathbb{R}$ then $xax^-1 \in H$ therefore $H \trianglelefteq G$.
Elements of $G/H = {H + (0, y): y \in \mathbb{R}}$.
Coset addition
$H = {(x, y): y = -x}$
For any $a \in H$ and $x \in G = \mathbb{R} \times \mathbb{R}$ then $xax^{-1} \in H$ therefore $H \trianglelefteq G$.
Elements of $G/H = {H + (0, y): y \in \mathbb{R}}$.
Coset addition
$H = {(x, y): y = 2x}$
Let $(\bar{x}, \bar{y}) \in H$ and $(u, v) \in \mathbb{R} \times \mathbb{R}$.
Then $(u, v)(\bar{x}, \bar{y})(u, v)^{-1} = (\bar{x}, \bar{y}) \in \mathbb{R} \times \mathbb{R}$, therefore $H \trianglelefteq G$.
Elements of $G/H = {H + (0, y): y \in \mathbb{R}}$.
Coset addition
If $x^2 \in H$ for every $x \in G$ then every element of $G/H$ is its own inverse.
Let there be a coset $Hx$, then $x^2 \in H$. So $\therefore x^2H = Hx^2 = H$. So $H$ is the identity coset.
$(Hx)(Hx) = Hx^2 = H$.
So every element of $G/H$ is its own inverse.
Likewise if every element of $G/H$ is its own inverse, then $(Hx)(Hx) = H \implies x^2 \in H$.
Let $m$ be a fixed integer. If $x^m \in H$ for every $x \in G$ then the order of every element in $G/H$ is a divisor of $m$.
Let there be an element $y \in G$ st. $y^m \in H$ where $m = qn$, therefore $(y^n)^q \in H$ where $ord(y) = n$. Then:
$$ (Hy)^n = (Hy) \cdots (Hy) = Hy^n = H $$
Conversely if the order of every element in $G/H$ is a divisor of $m$, then $x^m \in H$ for every $x \in G$.
This holds true because $ord(x) = n$, then $x^n = e = (x^n)^q = x^m$, where $m = qn$.
$$\therefore x^m \in H$$
Let $h = Hx$ then $ord(h) = n$ because $(Hx)^n = Hx^n = H$ because $x^n \in H$.
Suppose that for every $x \in G$, there is an integer $n$ st. $x^n \in H$.
Then every element of $G/H$ has a finite order. By previous exercise this is shown.
Every element of $G/H$ has a square root iff for every $x \in G$, there is some $y \in G$ st. $xy^2 \in H$.
$$xy^2 \in \implies xy^2 = h \text{ where } h \in H$$
$\therefore x = hy^{-2}$ but since $y \in G$ and $G$ is closed, there exists $\bar{y} \in G$ st. $\bar{y} = y^{-1}$ and $\therefore x = h\bar{y}^2$ and $x \in H\bar{y}^2$.
Theorem 5 also states:
iff $xy^2 \in H$ then $Hx = Hy^{-2} = (Hy)^{-2}$.
$G/H$ is cyclic iff there is an element $a \in G$ that $\forall x \in G$, $\exists$ integer $n$ st. $xa^n \in H$.
\begin{align*}
xa^n \in H \implies Hx &= Ha^{-n} \
&= (Ha)^{-n} = (Ha^{-1})^n
\end{align*}
Thus $G/H$ is cyclic since $(Ha^{-1})^n \in G/H$ because $a^{-1} \in G$.
$G$ is abelian, $H_p$ is the set of all $x \in G$ whose order is a power of $p$. Prove $H_p$ is a subgroup of $G$.
Property 1: closure
Let $x, y \in H_p$, then $ord(x) = p^k$ and $ord(y) = p^l$. That is, $x^{p^k} = e = y^{p^l}$.
Let $(xy)^{p^m} = e = x^{p^m} y^{p^m} \therefore m = lcm$ and $xy \in H_p$
Property 2: inverses
Let $x \in H_p$ and $e \in H_p$
\begin{align*}
x \cdot x^{-1} = e &= (x \cdot x^{-1})^{p^k} \
&= x^{p^k} (x^{-1})^{p^k} = (x^{-1})^{p^k} = e \
\therefore x^{-1} \in H_p
\end{align*}
Second part: prove that $G/H_p$ has no elements who order is a nonzero power of p.
Let $x \in G$ st $Hx \neq H_p$ and $ord(Hx) = p^k$.
Then $(Hx)^{p^k} = H_p$
\begin{align*}
\therefore h_1^{p^k} x^{p^k} &= h_2 \
x^{p^k} &= h_2 h_1^{-p^k}
\end{align*}
But $h_2 \in H_p$ and $h_1 \in H_p$
$$\therefore x^{p^k} = h \text{ where } h \in H_p$$
$$\therefore x^{p^k} \in H_p$$
But $x^{p^k} \in Hx \neq H_p$. Proof by contradiction.
If $G/H$ is abelian then:
$$HxHy = HyHx \text{ or } Hxy = Hyx$$
So $h_1xy = h_2yx$ where $h_1, h_2 \in H$
\begin{align*}
xy &= h_1^{-1} h_2 y x \
xyx^{-1} &= h_1^{-1} h_2 y \
xyx^{-1}y^{-1} &= h_1 h_2 \in H
\end{align*}
So all commutators of $G$ are in $H$ iff $G/H$ is abelian.
$H \trianglelefteq K \trianglelefteq G$ and $G/H$ is abelian. Prove $G/K$ and $K/H$ are both abelian.
From page 152, if $G/H$ is abelian, then it contains all the commutators of $G$.
Since $H \trianglelefteq K$, then:
$$Hxy = Hyx \text{ or } xy(xy)^{-1} \in H$$
Since all commutators are in $H$ and $H \trianglelefteq K$, then $G/H$ is abelian and so also $G/K$ because all commutators are also in $K$.
\begin{gather*}
K/H \text{ is abelian } \implies Hx, Hy \in K/H \
xyx^{-1}y^{-1} \in H \
Hxyx^{-1}y^{-1} = H \
Hxy = Hyx
\end{gather*}
So $K/H$ is abelian.
If every element of $G/H$ has finite order, and every element of $H$ has finite order, then every element of $G$ has finite order.
For every $h \in G/H$, $ord(h)$ is a divisor of $(G:H)$ by lagrange's theorem.
$$(G:H) = \frac{ord(G)}{ord(H)}$$
$$ord(G) = (G:H) ord(H)$$
But $ord(h)$ is a divisor of $(G:H)$. So:
$$ord(G) = (k \cdot ord(h)) ord(H)$$
If every element of $G/H$ has a square root, and every element of $H$ has a square root, then every element of $G$ has a square root. (Assume $G$ is abelian.)
Let $Hx \in G/H$ and $h \in H$.
If $x = y^2$ for some $y \in G$ and $h = \bar{h}^2$ for some $\bar{h} \in H$, then $hx = \bar{h}^2 y^2 = (\bar{h} y)^2$ since G is abelian.
$G/H$ and $H$ are p-groups $\implies \forall Hx \in G/H, (Hx)^{p^k} = H$
Because $H \trianglelefteq G$, $(Hx)^{p^k} = (Hx)\cdots(Hx) = Hx^{p^k}$, then:
$$x^{p^k} = h \in H$$
But,
\begin{align*}
h^{p^l} &= e \
(x^{p^k})^{lcm(l, k)} &= e^{lcm(l, k)} = e \
\therefore x^{p^{k\cdot lcm(l, k)}} &= e
\end{align*}
So every element of $G$ is a power of prime p.
Let $H$ be generated by ${h_1, \cdots, h_n}$ and let $G/H$ be generated by ${Ha_1, \cdots, Ha_m}$. Thus every element $x$ in $G$ can be written as a linear combination of $h_i$ and $a_j$.
For each element $a \in G$, the order of the element $Ha$ in $G/H$ is a divisor of the order of $a$ in $G$.
From Chapter 14, F1, if $f: G \rightarrow H$, then for each element $a \in G$, let $ord(a) = n$, then $a^n = e$ and $f(a^n) = (f(a))^n$, therefore the order of $f(a)$ is a divisor of the order of $a$ because $f(a^n) = f(e) = e_H$.
So therefore for each each element $a \in G$, let $ord(a) = n$, then $a^n = e$.
Then $(Ha)^n = He$ and so the order of $Ha$ in $G/H$ is a divisor of the order of $a$ in $G$.
If $(G:H) = m$, the order of every element of $G/H$ is a divisor of $m$.
$(G:H)$ is the order of $G/H$.
By theorem 5 (page 129): "the order of any element of a finite group divides the order of the group."
So if $(G:H) = m$, the order of every element of $G/H$ is a divisor of $m$.
If $(G:H) = p$ where $p$ is a prime, then the order of every element $a \notin H$ in $G$ is a multiple of $p$.
From theorem 5:
$$(G:H) = \frac{ord(G)}{ord(H)}$$
That is:
\begin{align*}
ord(G) &= (G:H) ord(H) \
&= p \cdot ord(H)
\end{align*}
Since the order of every element of G is a divisor of the order of G, then:
\begin{align*}
ord(a) = q \text{ and } ord(G) &= qn \
&= p \cdot ord(H)
\end{align*}
It follows that since $q | p ord(H)$ and $q \perp p$, then $q | ord(H)$ and so is a multiple of p.
If $G$ has a normal subgroup of index $p$, where $p$ is a prime, then $G$ has at least one element of order $p$.
$H \trianglelefteq G$ st $(G:H) = p$ where p is prime.
$$ord(G/H) = p$$
The order of $G/H$ is prime, thus it is cyclic.
Cauchy's theorem (page 131): "if $G$ is a finite group, and $p$ is a prime divisor of $|G|$, then $G$ has an element of order $p$."
Theorem 4 (page 129): "If $G$ is a group with a prime number $p$ of elements, then $G$ is a cyclic group. Furthermore, any element $a \neq e$ in $G$ is a generator of $G$."
So then $(G/H) \cong \mathbb{Z}_p$
If $(G:H) = m$, then $a^m \in H$ for every $a \in G$.
By Q2, $ord(Hx)$ is a divisor of m.
So $(Ha)^m = H$ but $H^m = H$ and $H$ is a normal subgroup of $G$, so $a^m \in H$.
In $\mathbb{Q}/\mathbb{Z}$, every element has finite order.
$$\mathbb{Q} = { p_1 / q_1 : p_1 q_1 = p_2 q_2 \forall p_1, p_2, q_1, q_2 \in \mathbb{Z}}$$
Where $(p_1, q_1) ~ (p_2, q_2)$ iff $p_1 q_1 = p_2 q_2$.
$$\mathbb{Q}/\mathbb{Z} = {m/n + \mathbb{Z}: m, n \in \mathbb{Z}}$$
Let $h \in \mathbb{Z}$, then $h^x \in \mathbb{Z}$ for any $x \in \mathbb{Z}$.
Then for any $g \in \mathbb{Q}/\mathbb{Z}$, $g^x$ is a coset of $m/n + \mathbb{Z}$
Therefore every element in $\mathbb{Q}/\mathbb{Z}$ has finite order.
For every $x \in G$, there is some integer $m$ such that $Cx = Ca^m$.
$$G/C = = { (Ca)^m : m \in \mathbb{Z} }$$
Now for $x \in G, Cx \in G/C$
$$\therefore \exists m : Cx = Ca^m$$
For every $x \in G$, there is some integer $m$ such that $x = ca^m$, where $c \in C$.
$$Cx = Ca^m \implies c_1 x = c_2 a^m \text{ where } c_1, c_2 \in C$$
\begin{align*}
c_1 x &= c_2 a^m \
&= c_1^{-1} c_2 a^m
\end{align*}
But $C$ is closed so $c_1^{-1} c_2 = c \in C$. So:
$$x = ca^m$$
For any two elements $x$ and $y$ in $G$, $xy = yx$.
\begin{align*}
x &= c_1 a^m \
y &= c_2 a^n \
xy &= c_1 a^m c_2 a^n
\end{align*}
But for any $c \in C$ and $x \in G$,
$$xc = cx$$
And $c_1, c_2 \in G$, so $c_1 c_2 = c_2 c_1$.
\begin{align*}
xy &= c_1 a^m c_2 a^n \
a^{-n} xy &= c_1 a^m c_2 \
c_2^{-1} a^{-n} xy &= c_1 a^m \
(a^n c_2)^{-1} xy &= c_1 a^m \
(a^n c_2)^{-1} x &= c_1 a^m y^{-1}
\end{align*}
But,
\begin{align*}
a^n c_2 &= c_2 a^n \
y^{-1} x &= c_1 a^m y^{-1} \
y^{-1} x &= xy^{-1} \
xy &= yx
\end{align*}
If $G/C$ is cyclic then:
$$x = ca^m \text{ for every } x \in G$$
And for any two elements in $G$, $xy = yx$.
Therefore $G$ is abelian.
Using the class equation to determine the size of the center.
Conjugancy class of $a$ is:
$$[a] = { xax^{-1} : x \in G }$$
The center of $G$ is:
$$C = { a \in G : xa = ax, \forall x \in G }$$
If $a \in C$ then for all $x \in G$:
\begin{align*}
xa &= ax \
xax^{-1} &= a
\end{align*}
This means the conjugancy class of $a$ contains $a$ (and no other element).
Let $c$ be the order of $C$. Then $|G| = c + k_s + k_s + k_{s + 1} + \cdots + k_t$, where $k_s, \cdots, k_t$ are the sizes of all the distinct conjugacy classes of elements $x \notin C$.
$$C = { a \in G : xax^{-1} = a, \forall x \in G }$$
If $a \in C$ then $xax^{-1} = a$ for all $$x \in G$ and $[a] = {a}$.
So $c = k_1 + \cdots + k_{s - 1}$ and $|G| = c + k_s + \cdots + k_t$ where $k_s, \cdots, k_t$ are sizes of distinct conjugacy classes of elements $a \notin C$.
For each $i \in {s, s + 1, \cdots, t}$, $k_t$ is equal to a power of $p$.
Chapter 13, I6 states "the size of every conjugancy class is a factor of |G|". $|G| = p^k$ so $|S_i| = k_i$ must equal some factor of $p^k$, that is, there is some $p^m$ which divides $p^k$.
See this video at 17:20 for the proof.
Explain why $c$ is a multiple of $p$.
From orbit-stabilizer $k_i = \frac{|G|}{|C_x|}$ where $|G|$ has a prime divisor $p$.
But $k = c + k_s + \dots + k_t$ where $k$ and all $k_i$ are factors of $p$, so $c$ is a factor of $p$ also.
If $|G| = p^2$, $G$ must be abelian.
By lagrange's theorem $|C| | |G|$.
Possibilities are ${1, p, p^2}$.
$|C| \neq 1$ because center is non-trivial.
If $|C| = p$, then $G/C$ has $p$ cosets, therefore $G/C$ is cyclic and hence abelian (from part F).
Else $|C| = p^2$ means $C$ is entire group and abelian.
Any group of size $p^2$ is isomorphic to $\mathbb{Z}_{p^2}$ or $\mathbb{Z}_p \times \mathbb{Z}_p$.
To see why, if there is an element $ = \mathbb{Z}{p^2}$ then the group is isomorphic to $\mathbb{Z}{p^2}$.
If not then by lagrange's theorem, the subgroup must have order $p$, in which case the group is isomorphic $\mathbb{Z}_p \times \mathbb{Z}_p$ by the mapping:
$$f(x) : G \rightarrow \mathbb{Z}_p \times \mathbb{Z}_p$$
By $f(ab) = (a, b)$.
See also Cayley's theorem on page 96.
If $ord(a) = tp$ where $a \in G$, what element of $G$ has order $p$?
$$ord(a) = tp \implies a^{tp} = e = (a^t)^p$$
Therefore $ord(a^t) = p$
Now $ord(a)$ is not a multiple of $p$. Then $G/$ is a group with fewer than $k$ elements and its order is a multiple of $p$.
$|G| = k = np$ where $p$ is prime but $ord(a)$ is not a multiple of $p$.
By lagrange's theorem $ord(a)$ must divide $|G|$ since $$ is a subgroup of $G$.
$ord(a) | k$ or $ord(a) | np$, but since $ord(a) \perp p$ then $ord(a) | n$.
The order of $G/$ is the same as the number of cosets of $$.
\begin{align*}
ord(G/) &= (G : ) \
&= \frac{ord(G)}{ord(a)}
\end{align*}
Since $ord(a)$ is not a multiple of $p$, but $|G|$ is, then $ord(G/)$ is a multiple of $p$.
Since $ord(G/)$ is a multiple of $p$, by Cauchy's theorem, $p$ is a prime divisor of the group, then $G/$ has an element of order $p$.
By E1, $G$ has an elemtn of order $p$, by an isomorphism from $f(a) = \bar{a}$.