Elementary definitions
Definition (Binary operation)
Let \(G\) be a non-empty set. A binary operation is a map \(G \times G \to G\) with \((g_{1},g_{2}) \mapsto g_{3}\).
Definition (Group)
A group \(G\) is a non-empty set together with a binary operation \(\ast: G \times G \to G\) such that:
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\(\ast\) is associative:
\[\forall g_{1},g_{2},g_{3} \in G \quad (g_{1} \ast g_{2}) \ast g_{3} = g_{1} \ast (g_{2} \ast g_{3});\] - there exists an identity element \(e \in G\) such that \(e \ast g = g\ast e = g\) for every \(g \in G\). We often write \(1_{G}\) to denote the identity element of the group \(G\);
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every element has an inverse, that is,
\[\forall g \in G \quad \exists h \in G \quad (g \ast h = h \ast g = e).\]
Definition (Abelian Group)
If \((G, \ast)\) is a group which moreover satisfies that for all \(g_{1},g_{2} \in G\), \(g_{1} \ast g_{2} = g_{2} \ast g_{1}\), we say that \(G\) is commutative or abelian.
Definition (Monoid)
A monoid is a more general structure, given by a set and an associative binary operation with an identity element, but where not every element necessarily has an inverse.
Properties of groups
Proposition (Uniqueness of the identity)
If \(e, e'\) are two identity elements, \(e' = e' \ast e = e\).
Proposition (Uniqueness of the inverse)
Let \(g \in G\) and let \(h, h'\) be inverses of \(g\). Then
\[h' = e \ast h' = (h \ast g) \ast h' = h \ast (g \ast h') = h \ast e = h.\]Notation (Inverse)
If \((G, \ast)\) is a group and \(g \in G\), we denote by \(g^{-1}\) the inverse of \(g\).
Note (Properties of the inverse)
- \((g^{-1})^{-1} = g\) for every \(g \in G\)
- \((g \ast h)^{-1} = h ^{-1} \ast g ^{-1}\) for all \(g,h \in G\).
Proposition (Cancellation laws)
If \(g, h_{1}, h_{2} \in G\), then
- \(g \ast h_{1} = g \ast h_{2} \implies h_{1}=h_{2}\),
- \(h_{1} \ast g = h_{2} \ast g \implies h_{1} = h_{2}\).
Examples of groups
Some examples are very important within the theory
Example (Trivial group)
Let \(G = \{ e \}\) be such that \(e \ast e = e\). \((G, \ast)\) is a group and is called the trivial group.
Example (Additive groups)
\((\mathbb{Z}, + )\) is a group with identity element \(0\) and in which the inverse of \(z \in Z\) is \(-z\). \((\mathbb{N}, +)\) is not a group but it is a monoid, whereas \((\mathbb{N}^{\ast},+)\) is not a monoid. \((\mathbb{R},+), (\mathbb{Q},+), (\mathbb{C}, +), (\mathbb{Z},+)\) are all abelian groups.
Example (Multiplicative groups)
\((\mathbb{R}^{\ast}, \cdot), (\mathbb{Q}^{\ast}, \cdot), (\mathbb{C}^{\ast}, \cdot)\) are abelian groups with identity \(1\) and where the inverse of \(x\) is \(\frac{1}{x}\). In general, if \(\mathbb{F}\) is a field, \((\mathbb{F}^{\ast}, \cdot)\) is a group. \((\mathbb{Z}^{\ast}, \cdot)\) is not a group because not every element has an inverse.
Example (Linear groups)
Let \(n \in \mathbb{N}^{\ast}\) and let \(\mathbb{F}\) be a field. We define
\[GL_{n}(\mathbb{F}) = \{\text{invertible }n \times n \text{ matrices with entries in }\mathbb{F} \}.\]\((GL_{n}(\mathbb{F}), \cdot)\), where \(\cdot\) is the standard multiplication of matrices. This group is called the linear group over \(\mathbb{F}\).
Example (Groups over congruences)
Let \(m>0\) and consider \(\mathbb{Z}_{m} = \mathbb{Z}/{\equiv_{m}}\) (\(\mathbb{Z}\) quotiented by the relation of congruence modulo \(m\)), i.e., \(\mathbb{Z}_{m} = \{ [0], [1],\dots, [m-1] \}\). Define the operation \(\hat{+}\) on \(\mathbb{Z}_{m}\) such that \([a]\hat{+}[b] = [a+b]\) (we must show that it is well defined). \((\mathbb{Z}_{m}, \hat{+})\) is a group with identity element \([0]\). If \([n] \in \mathbb{Z}_{m}\), the inverse is \([-n]\). Let us consider the example with \(m=2\).
For \(\mathbb{Z}_{2}\), we have \(\mathbb{Z}_{2} = \{ [0]. [1] \}\). Note the operation table
| \(\hat{+}\) | \([0]\) | \([1]\) |
|---|---|---|
| \([0]\) | \([0]\) | \([1]\) |
| \([1]\) | \([1]\) | \([0]\) |
Let us now define on \(\mathbb{Z}_{m}^{\ast} = \mathbb{Z}_{m} \setminus \{ [0] \}\). Let us define the operation \(\hat{\cdot}\) as
\[[a] \hat{\cdot} [b] = [a \cdot b]\]We want to construct a multiplicative group with this new operation. For which cases of \(m\) is this well defined and do inverses exist? For composite \(m\), multiplication is not well defined on \(\mathbb{Z}_{m}^{\ast}\) because we may take elements of certain equivalence classes whose product has result \([0]\).
For primes, the operation is well defined. Let \(n\) be prime and suppose that there exist \([a], [b] \in \mathbb{Z}_{n}^{\ast}\) such that \([a] \hat{\cdot} [b] = [a \cdot b] = [0]\). This implies that \(n \mid a \cdot b\), but since \(n\) is prime, \(n \mid a\) or \(n \mid b\), so that \([a] = [0]\) or \([b]=[0]\), a contradiction since \([0] \not\in \mathbb{Z}_{n}^{\ast}\).
We shall prove that \((\mathbb{Z}_{n}^{\ast}, \hat{\cdot})\) is a group for \(n\) prime. Note that \(\mathbb{Z}_{n}^{\ast} = \{ [1],\dots,[n-1] \}\). Let \(r \in \{ 1,\dots,n-1 \}\). Since \(n\) is prime, we have that \(\operatorname{GCD}(r,n) = 1\). Hence there exist integers \(q,s\) such that \(qr+sn = 1\). This implies that
\[[q \cdot r] = [1] \implies [q] \hat{\cdot} [r] = [1].\]We may suppose that \(q \in \{ 1,\dots,n-1 \}\), since otherwise we take the representative of its equivalence class. Thus \([q]\) is the inverse of \([r]\). So \((\mathbb{Z}_{n}^{\ast}. \hat{\cdot})\) has inverses for every element, and therefore it is a group (the other properties are trivial).
Note (Field \(\mathbb{Z}_p\))
It is easy to verify that if \(p\) is prime \((\mathbb{Z}_{p}, \hat{+}, \hat{\cdot}, [0], [1])\) is a field.
Example (Groups of symmetries of a square)
Consider a square with its vertices numbered. In how many ways can we move its vertices so that the square still occupies the same space? We can perform four rotations (\(0^{\circ}, 90^{\circ}, 180^{\circ}, 270^{\circ}\)) anticlockwise, (\(r_{0},r_{90},r_{180},r_{270}\)). Moreover, we can perform four reflections: vertical (\(s_{1}\)), horizontal (\(s_{2}\)), diagonal starting bottom-left (\(s_{3}\)) and diagonal starting top-left (\(s_{4}\)).
We call \(D_{4}= \{ r_{0},r_{90},r_{180},r_{270}, s_{1},s_{2},s_{3},s_{4} \}\). We define an operation \(\circ\) on \(D_{4}\), as the “composition of movements”, e.g., \(r_{90} \circ s_{2}\).
Exercise (Dihedral group \(D_4\))
\((D_{4},\circ)\) is a group but is not commutative.
In general \(D_{n}\) is the group of symmetries of a regular polygon with \(n\) sides. It is made up of \(r_{0},\dots,r_{n-1}\) rotations, where \(r_{k}\) rotates the figure by an angle of \(\frac{2\pi k}{n}\) and \(s_{0}, \dots, s_{n-1}\) are \(n\) reflections about the lines \(\theta = \frac{\pi k}{n}\). \(D_{n}\) has \(2n\) elements, and is called the dihedral group.
Example (Group of permutations of \(n\) elements)
A permutation of \(n\) elements is a bijective function
\[\begin{aligned} \sigma:\{ 1,\dots,n \} &\to \{ 1,\dots,n \}, \text{ such that} \\ 1 &\mapsto \sigma(1)\\ & \quad\vdots \\ n &\mapsto \sigma(n). \end{aligned}\]We define \(S_{n} = \{ \sigma: \sigma \text{ is a permutation of }n \text{ elements}\}\). \(S_{n}\) is a group under composition of functions (exercise).
Notation (Permutations)
\(\sigma: \begin{pmatrix}1 & 2 & 3 \\ 2 & 3 & 1\end{pmatrix}\) represents \(\sigma:\{ 1,2,3 \} \to \{ 1,2,3 \}\), with \(\sigma(1)=2, \sigma(2)=3,\sigma(3)=1\). Its inverse is \(\sigma ^{-1}:\begin{pmatrix}1 & 2 & 3 \\ 3 & 1 & 2\end{pmatrix}\).
Example (Product of groups)
Let \((G_{1}, \ast_{1})\) and \((G_{2}, \ast_{2})\) be two groups. Consider the Cartesian product \(G_{1} \times G_{2}\). Define the operation \(\ast\) on \(G_{1} \times G_{2}\) such that
\[(g_{1}, g_{2}) \ast (h_{1},h_{2}) := (g_{1} \ast_{1} h_{1}, g_{2} \ast_{2} h_{2}) \in G_{1} \times G_{2}.\]Then, \((G_{1} \times G_{2}, \ast)\) is a group with identity element \((1_{G_{1}}, 1_{G_{2}})\). If \((g_{1},g_{2}) \in G_{1} \times G_{2}\), then \((g_{1}, g_{2})^{-1} = (g_{1}^{-1}, g_{2}^{-1})\). This can be generalised to the product of \(n\) groups.
Iterated operations in groups
If \((G, \ast)\) is a group we denote by \(g^{n} = \underbrace{ g \ast g \ast \dots \ast g }_{ n \text{ times} }\),, \(g^{0} = 1_{G}\), \(g^{-n}:= (g^{-1})^{n}\).
Proposition (Properties of exponentiation)
Let \((G, \ast)\) be a group, \(g,h \in G\). Then
- \[g^{-n} = (g^{n}) ^{-1}\]
- \[g^{n+m} = g^{n} \ast g^{m}\]
- If \(g \ast h = h \ast g\), then \((g \ast h)^{m} = g^{m} \ast h^{m}\).
Subgroups
Let \((G, \ast)\) be a group and \(H \subseteq G\). We say that \(H\) is a subgroup of \(G\) if \((H, \ast \mid_{H})\) is a group. In that case we write \(H \leq G\)
Note (Subgroup verification)
To verify that \(H \leq G\), it suffices to check (1) that \(h_{1}\ast h_{2} \in H \quad \forall h_{1}, h_{2} \in H\), (2) that \(1_{G} \in H\) and (3) that, if \(h \in H\) then \(h^{-1} \in H\).
Theorem (Characterisation of a subgroup)
Let \(\emptyset \neq H \subseteq G\). Then \(H \leq G\) if and only if
\[\forall x,y \in H \quad (x \ast y^{-1} \in H).\]Proof: \((\implies)\) is trivial. Let us prove \((\impliedby)\). If \(x \in H\), then \(x \ast x ^{-1} = 1_{G} \in H\). On the other hand, since \(1_{G} \in H\), \(1_{G} \ast x^{-1} = x^{-1} \in H\). Finally, if \(x,y \in H\), since \(y^{-1} \in H\), then \(x \ast (y ^{-1}) ^{-1} = x \ast y \in H\). Therefore, \(H\) is a subgroup.
Lemma (The intersection of subgroups is a subgroup)
Let \((G, \ast)\) and let \(\{ H_{i}:i \in I \}\) be a family of subgroups of \(G\). Then \(H:=\bigcap_{i \in I} H_{i} \leq G_{i}\).
Proof: Note that \(H\neq \emptyset\), since \(1_{G} \in H_{i}\) for every \(i\). Let \(x,y \in H\). Since \(x,y \in H_{i}\) for every \(i \in I\). Then \(x \ast y^{-1} \in H_{i}\) for every \(i\). Hence \(x \ast y ^{-1} \in H\).
Definition (Group generated by a subset)
Let \((G, \ast)\) be a group and \(S \subseteq G\). Define
\[\langle S \rangle := \bigcap \{ H:H\leq G \quad \land \quad S \subseteq H\}\]as the group generated by \(S\).
Theorem (The generated group is the smallest group containing the set)
\(\langle S \rangle\) is the smallest group containing \(S\), i.e., if \(K\leq G\) is such that \(S \subseteq K\), then \(\langle S \rangle \subseteq K\).
Proof: If \(K \leq G\) and \(S \subseteq K\), then \(K \in \{ H \leq G : S \subseteq H \}\), so that \(\bigcap \{ H:H\leq G \quad \land \quad S \subseteq H\} = \langle S \rangle \subseteq K\).
Theorem (Characterisation of the generated group)
If \((G, \ast)\) is a group and \(S \neq \emptyset\), then
\[\langle S \rangle = \{ x_{1}^{\alpha_{1}} \ast \dots \ast x_{n}^{\alpha_{n}}: n \in \mathbb{N}^{\ast}, x_{i} \in S, \alpha_{i} \in \{ -1,1\} \}.\]Proof: Let \(H = \{ x_{1}^{\alpha_{1}} \ast \dots \ast x_{n}^{\alpha_{n}}: n \in \mathbb{N}^{\ast}, x_{i} \in S, \alpha_{i} \in \{ -1,1\} \}\). Note that \(H \neq \emptyset\). Let \(x,y \in H\). Then there exist \(k, \ell \in \mathbb{N}^{\ast}; \{ \alpha_{i} \}_{i=1}^{k}, \{ \beta_{i} \}_{i=1}^{\ell}\) with \(\alpha_{i}, \beta_{i} \in \{ -1,1 \}\) for every \(i\), and \(x_{1},\dots,x_{k},y_{1},\dots y_{\ell} \in S\) such that
\[x = x_{1}^{\alpha_{1}} \ast \dots \ast x_{k}^{\alpha_{k}}, y = y_{1}^{\beta_{1}}\dots y_{\ell}^{\beta_{\ell}}.\]Note that \(x \ast y^{-1} = x_{1}^{\alpha_{1}} \ast \dots \ast x_{k}^{\alpha_{k}} \ast y_{1}^{-\beta_{1}}\dots y_{\ell}^{\beta_{-\ell}} \in H\). Thus \(H \leq G\) and therefore \(\langle S \rangle \subseteq H\). On the other hand, since \(\langle S \rangle\) is a group, \(H \subseteq \langle S \rangle\).
Definition (Generated group)
If \((G, \ast)\) is a group and \(S \subseteq G\) is such that \(\langle S \rangle = G\), we say that \(G\) is generated by \(S\). If \(S\) is finite, we say that \(G\) is finitely generated.
Examples (of generated groups)
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Let \((G, \ast)\) be a group. Then
\[\begin{aligned} \langle \emptyset \rangle &= \bigcap \{ H: H\leq G \} \\ &= \{ 1_{G} \}. \end{aligned}\] - If \((G, \ast)\) is a group and \(S \leq G\), \(\langle S \rangle = \langle S \rangle\).
- If \(S = \{ i \} \subseteq (\mathbb{C}^{\ast}, \cdot)\), then \(\langle S \rangle = \{ i,-1,1,-i \}\).
- If \(S = \{ 2 \} \subseteq (\mathbb{Z},+)\), \(\langle S \rangle = \{ 2k:k \in \mathbb{Z} \}\).
- If \(S=\{ 12,-8 \} \subseteq (\mathbb{Z}, +)\), then \(\langle S \rangle = \{ 12 a + (-8)b: a,b \in \mathbb{Z} \} = \{ 4k: k \in \mathbb{Z} \}\) .
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In general, if \(a \in (G, \ast)\),
\[\begin{aligned} \langle a \rangle &= \bigcap \{ H: H<a \quad\land \quad a \in H \}\\ &= \{ a^{n}: n \in \mathbb{Z} \}. \end{aligned}\]
Centre and centralizer
Definition (centre of \(G\))
Let \((G, \ast)\) be a group. Define
\[Z(G) = \{ g \in G: \forall h \in G (gh = hg)\}\]Note (Properties of the centre)
- \(e \in Z(G)\),
- \(G\) is abelian if and only if \(Z(G) = G\).
- If \(g \in Z(G)\), then for every \(h \in G\), \(h = g \ast h \ast g^{-1}\).
Theorem (The centre is a subgroup)
If \(G\) is a group, then \(Z(G) \leq G\) and \(Z(G)\) is commutative.
Definition (Centralizer)
Let \(G\) be a group and \(a \in G\). Define the centralizer as
\[C(a):= \{ g \in G: a \ast g = g \ast a \}\]Exercise (The centralizer is a subgroup)
\(C(a) \leq G\).
It can be proved that \(Z(G) = \bigcap_{a \in G} C(a)\).