Examples of Groups
- Integers: $(\mathbb{Z},+)$. Remember that this does not form a group under multiplication (think about inverses)
- Integers modulo $n$: $(\mathbb{Z}/n\mathbb{Z}, +)$. There is an associated multiplicative group called the group of units $(U(n), \times)$
- Symmetries of a (regular) polygon.
We will add a new one to the list:
$\mathbb{C}^\times$: Consider the non-zero complex numbers: $$\mathbb{C}^\times = \left \{ a + bi \mid a,b \in \mathbb{R},a,b \text{ not simultaneously 0} \right \}$$
$\mathbb{C}^\times$ forms a multiplicative group.
Proof Sketch
Check closure, associativity, inverse, and identity. As a hint, associativity comes down to checking that
$$(a+bi)[(c+di)(e+fi)] \stackrel{?}{=} [(a+bi)(c+di)](e+fi)$$Similarly, inverse comes down to solving
$$z \cdot \alpha = 1 \Rightarrow \alpha = \frac{1}{a + bi} = \frac{a}{a^2 + b^2} - i \frac{b}{a^2 + b^2}$$Is this group commutative?
Subgroups
Let $(G,\circ)$ be a group. A subgroup is a subset $H \subseteq G$ such that $(H,\circ)$ is also a group with the same operation. In this case, we write $H \leq G$.
Let $\emptyset \neq H \subseteq G$. Then $(H,\circ)$ is a subgroup if and only if for all $g,h \in H$, $g h^{-1} \in H$.
Proof Sketch
The forwards direction is clear from the axioms. In the converse direction, see first that $g \in H$ implies that $g g^{-1} = e_{G} \in H$, so identity holds. Next, we have $e_{G} g^{-1} = g^{-1} \in H$ implies that inverse holds. Since associativity holds for the entire group, $H$ inherits the property, so it remains to check closure. For any $h_{1}, h_{2} \in H$, consider the product
$$h_{1} h_{2} = h_{1} (h_{2}^{-1})^{-1} \in H$$Find all subgroups of the Quaternion group $Q_{8}$
Subgroup lattice:
Let $gAg^{-1} = \{ gag^{-1} \mid a \in A \}$. Define the normalizer of $A$ in a group $G$ as $$N_G(A) = \{ g \in G \mid gAg^{-1} = A \}.$$ Show this is a subgroup of $G$.
Proof
We'll show it is non-empty. Let $e \in G$ be the identity. Then
$$eAe = A,$$so $e \in N_G(A)$. Next, we show the subgroup criterion. Let $g, h \in N_G(A)$. Then
$$gAg^{-1} = A, \qquad hAh^{-1} = A.$$We want to show
$$gh^{-1}A(gh^{-1})^{-1} = A.$$First, see
$$gh^{-1}A(gh^{-1})^{-1} = gh^{-1}A\,hg^{-1}.$$Claim. If $h \in N_G(A)$, then $h^{-1} \in N_G(A)$.
Why? $h \in N_G(A) \implies hAh^{-1} = A \implies \forall a \in A,\ hah^{-1} = b \in A$ (also $\forall b,\ \exists a$).
$$\Updownarrow$$ $$\forall a \in A,\quad a = h^{-1}bh, \qquad h \in N_G(A).$$
Using the claim,
$$gh^{-1}A\,hg^{-1} = gAg^{-1} = A$$Cyclic groups
A group $(G, \circ)$ is cyclic if $\exists g \in G$ such that $\forall h \in G,\ \exists n$ such that $h = g^n$. Write $G = \langle g \rangle$.
- $\mathbb{Z} = \langle 1 \rangle$.
- $\mathbb{Z}/n\mathbb{Z} = \langle \overline{1} \rangle$.
If $(G, \circ)$ is a group, then $\forall a \in G$, $\langle a \rangle \leq G$ forms a subgroup, called the cyclic subgroup.
- $2 \in \mathbb{Z}$, $\langle 2 \rangle \leq \mathbb{Z}$.
- $i \in Q_8$, $\langle i \rangle = \{i, -1, -i, 1\} \leq Q_8$.
The order of $a \in G$ is the size of $\langle a \rangle$. If $\langle a \rangle = G$, $a$ is a generator.
What are the generators of $\mathbb{Z}/n\mathbb{Z}$?
$m \in \mathbb{Z}/n\mathbb{Z}$ such that $\gcd(m, n) = 1$.
How many numbers in $\{0, \dots, n-1\}$ have $\gcd = 1$?
How to find an order $k$ subgroup of $\mathbb{Z}/n\mathbb{Z}$?
If $b = a^{\ell}$, then $|b| = \dfrac{n}{d}$ with $d = \gcd(\ell, n)$.
Find the generators of the subgroup of order $7$ in $\mathbb{Z}/42\mathbb{Z}$.
Problems
$(\mathbb{Z}, +)$. Pick $n \in \mathbb{N}$, consider the set $$n\mathbb{Z} = {\dots, -2n, -n, 0, n, 2n, \dots}.$$ This is a subgroup.
Why?
Nonempty, so use the subgroup criterion: let $z_1, z_2 \in \mathbb{Z}$, consider $nz_1, nz_2 \in n\mathbb{Z}$. WTS: $nz_1 - nz_2 \in n\mathbb{Z}$. Surely, we have that $$nz_1 - nz_2 = n(z_1 - z_2) \in n\mathbb{Z},$$ as needed. $\square$
Let $(G, \circ)$ be a group, and let $a \in G$ arbitrary. The set $$\langle a \rangle = { a^k \mid k \in \mathbb{Z} } \leq G$$ forms a subgroup, called the cyclic subgroup generated by $a \in G$.
Why?
$\langle a \rangle \neq \emptyset$: $0 \in \mathbb{Z}$, so $a^0 = e \in \langle a \rangle$ ✓. Subgp. Crit.: Let $\alpha, \beta \in \langle a \rangle$. WTS $\alpha\beta^{-1} \in \langle a \rangle$. By definition, $\exists, n, m \in \mathbb{Z}$ s.t. $$\alpha \cdot \beta^{-1} = a^n \cdot (a^m)^{-1} = a^n \cdot a^{-m} = a^{n-m} \in \langle a \rangle.$$
Claim. $(\langle a \rangle, \circ)$ is abelian.
Why?
WTS $\forall \alpha, \beta \in \langle a \rangle$, $\alpha\beta = \beta \cdot \alpha$. Remember $\exists, n, m \in \mathbb{Z}$ s.t. $\alpha = a^n$, $\beta = a^m$, so $$\alpha\beta = a^n \cdot a^m = a^{n+m} = a^m \cdot a^n = \beta \cdot \alpha.$$
Claim. $(\langle a \rangle, \circ)$ is the smallest subgroup of $G$ containing $a \in G$.
Why?
Let $H \leq G$ with $a \in H$. By induction (+ closure) $a^n \in H$ $\forall, n \in \mathbb{Z}_{\geq 0}$, and by inverse $a^{-n} \in H$. Thus $\langle a \rangle \subseteq H$. $\square$
Claim. Subgroups of cyclic groups are cyclic.
Proof
Let $G = \langle a \rangle$ be a cyclic group, and let $H \leq G$ be a subgroup.
Case 1: $H = {e}$. Trivial.
Case 2: If $H \neq {e}$, $\exists, h \in H$ s.t. $h = a^n$ for $n > 0$ (if not, pick its inverse). Then consider the set $$S = { n \in \mathbb{Z}_{>0} \mid a^n \in H } \neq \emptyset \quad \leadsto \quad \exists \text{ least element by Well Ordering!}$$ Say $m$ is that least element. Call this element $a^m \in H$. We will show $H = \langle a^m \rangle$. Let $g \in H$, then $g = a^k$ for some $k \in \mathbb{Z}$. By the Division Algorithm $\exists, q, r \in \mathbb{Z}$ w/ $$k = qm + r \qquad \text{w/ } 0 \leq r < m,$$ $$H \ni g = a^k = (a^m)^q \cdot a^r.$$
If $r \neq 0$, then $g \cdot a^{-qm} = a^r \in H$, and $0 \leq r < m$ contradicts minimality of $m$. $\square$
True or False: All generators of $\mathbb{Z}_{60}$ are prime.
Solution
What does it mean to generate $\mathbb{Z}_{60}$? $\langle a \rangle = \mathbb{Z}_{60} \implies \underbrace{a + \cdots + a}_{n} = 1$ first when $n = 60$. This is the same as saying $$\gcd(a, 60) = 1.$$ Thus, we ask: is there an $n \in {1, \dots, 60}$ s.t. $\gcd(n, 60) = 1$ and $n$ not prime? $\boxed{49}$
True or False: If $a, b \in G$ have finite order, then $ab$ has finite order.
Solution
FALSE: $$\begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix}, \quad \begin{pmatrix} 0 & -1 \\ 1 & -1 \end{pmatrix} \in GL_2(\mathbb{R})$$
$|a| = |gag^{-1}|$ $\forall, g \in G$.
Let $G = \langle a \rangle$ be a cyclic group. Show that $\langle a^m \rangle \cap \langle a^n \rangle$ is cyclic. What's the generator?
Proof
If $\langle a^m \rangle \cap \langle a^n \rangle \ni b$, then $\exists, r_1, r_2 \in \mathbb{Z}_{>0}$ s.t. $$a^{r_1 m} = a^{r_2 n} = b \quad \leadsto \quad \text{like earlier, pick minimal such!}$$ This is the $\operatorname{lcm}(n, m)$.