# Review of Group Theory

Very likely you already know the basic properties of groups, rings and fields. Some of this Chapter is therefore review. We will be a little inconsistent in that we will develop group theory from scratch, but later when we begin the group representation theory, we will assume some properties of vector spaces and fields to be known to the reader. To accomplish a speedy development, we will leave many details to the reader in Exercises.

We will use the notation $X \subseteq Y$ to mean that $X$ is a subset of $Y$ that might equal $Y$, and $X \subsetneq Y$ if $X$ is not allowed to equal $Y$. Strictly speaking the notation $X \subset Y$ is equivalent to $X \subseteq Y$.

If $X$ is a set, we will denote by $|X|$ the cardinality of $X$. If $X$ is finite, this is just the number of elements of $X$. This notation is commonly used in group theory and combinatorics, but in other branches of mathematics it is not universally standard. If $G$ is a group, its cardinality $|G|$ is called its order, though this term is not usually used for sets that are not groups, and the term "order'' has other common meanings in mathematics.

A group is a set $G$ together with a map $\mu : G \times G \longrightarrow G$ (called the group law) and a distinguished element $1 \in G$ called the unit or identity element that satisfy certain axioms that we will state momentarily. We regard $\mu$ as a multiplication and denote $\mu (x, y)$ by $x \cdot y$ or $x y$. The axioms are:

• We have $1 \cdot x = x \cdot 1 = x$ for $x \in G$.
• We have $x \cdot (y \cdot z) = (x \cdot y) \cdot z$ if $x, y, z \in G$.
• If $x \in G$ there is an element $x^{- 1}$ such that $x \cdot x^{- 1} = x^{- 1} \cdot x = 1$. The map $x \longmapsto x^{- 1}$ from $G \longrightarrow G$ is called the inverse map.
It is sometimes useful to weaken the axioms a little and only assume the first two; in that case $G$ is called a monoid. Thus in a monoid there is a unit, and the multiplication is associative, but inverses are not assumed to exist. Examples of monoids are $\mathbb{Z}$ with respect to multiplication, or the set of nonnegative integers with respect to addition.

We may denote $1$ by $1_G$ if there are several groups floating around and it is necessary to distinguish them. This notation has a drawback since occasionally if $X$ is a set we will use $1_X$ to denote the identity map $X \longrightarrow X$.

Exercise 1.1.1: Prove that if $x, y$ are elements of a group, then $(x y)^{- 1} = y^{- 1} x^{- 1}$.

If $x y = y x$ for $x, y \in G$ then the group $G$ is called Abelian. For Abelian groups it is sometimes useful to use additive notation: thus we denote the identity element $0 \in G$ and denote the group law $m (x, y) = x + y$ instead of $x \cdot y$. However we will use multiplicative notation most of the time.

If $G$ is a group and $H$ is a subset of $G$ such that $1_G \in H$, and whenever $x, y \in H$ we have $x \cdot y, x^{- 1} \in H$, then $H$ is called a subgroup of $G$. We express this assumption by saying that $H$ is closed under multiplication and under the inverse map. We call $H$ a proper subgroup if it is subgroup that is a proper subset, that is, $H \neq G$.

If $G$ and $H$ are groups, and $\phi : G \longrightarrow H$ is a map, we call $\phi$ a homomorphism if $\phi (x \cdot y) = \phi (x) \cdot \phi (y)$. We express this identity by saying that $\phi$ "preserves'' the group law. It is easily deduced that $\phi (1_G) = 1_H$ and $\phi (x^{- 1}) = \phi (x)^{- 1}$ for $x \in G$. The image of $\phi$ is $\phi (G) =\{\phi (g) |g \in G\}$, and the kernel $\ker (\phi) =\{g \in G| \phi (g) = 1\}$. We will also sometimes denote the image $\phi (G) = \operatorname{im} (\phi)$.

Exercise 1.1.2: Prove that if $\phi : G \longrightarrow H$ is a homomorphism, then $\operatorname{im} (\phi)$ is a subgroup of $H$, and $\ker (\phi)$ is a subgroup of $G$. Prove that $\phi$ is surjective if and only if $\operatorname{im} (\phi) = H$ and injective if and only if $\ker (\phi) = 1$. (Here $1$ means the subgroup $\{1\}$ of $G$. Of course the surjectivity statement is obvious and is included only to point up a certain duality between image and kernel; but give details for the injectivity claim.)

If $X$ is a set, an equivalence relation on $X$ is a relation $\sim$ such that $x \sim y$ is defined to be true or false for $x$ and $y$ in $X$; the relation is assumed to satisfy the following axioms.

• Reflexive: $x \sim x$ for $x \in X$;
• Symmetric: $x \sim y$ if and only if $y \sim x$;
• Transitive: $x \sim y$ and $y \sim z$ implies $x \sim z$.
If $x \in X$, then the equivalence class $c (x) =\{y \in X| y \sim x\}$. When working with an equivalence relation, we will say $x$ is equivalent to $y$ if $x \sim y$.

Exercise 1.1.3: If $x, y \in X$, prove that $c (x)$ and $c (y)$ are either disjoint or equal. Moreover $x \in c (x)$ so $X = \bigcup_{x \in X} c (x)$. Thus $X$ is the disjoint union of the equivalence classes.

Exercise 1.1.4: Let $G$ be a group and $H$ a subgroup. Define a relation $\sim$ on $G$ by $x \sim y$ if $x^{- 1} y \in H$. Prove that $\sim$ is an equivalence relation.

If $X$ and $Y$ are subsets of a group $G$, we will use the notation $X Y$ to denote the set $\{x y| x \in X, y \in Y\}$, and $X^{- 1} =\{x^{- 1} |x \in X\}$. The identity $(X Y) Z = X (Y Z)$ for three subsets $X, Y, Z$ is an immediate consequence of the associative law, and we will use the notation $X Y Z =\{x y z| x \in X, y \in Y, z \in Z\}$ for this triple product, and similarly for products of several subsets. We will also write (for an element $x$ and a subset $Y$) $x Y =\{x y| y \in Y\}$ and other obvious notations.

Exercise 1.1.5: Let $G$ be a group and $H$ a subgroup. Let $\sim$ denote the equivalence relation of Exercise 1.1.4. Show that the equivalence class of $x$ is the left coset $x H$.

Proposition 1.1.1: Let $G$ be a group and let $H$ be a subgroup. The cardinality of the set of left cosets $x H$ equals the cardinality of the set of right cosets $H x$.

Proof. (Click to Expand/Collapse)

We have $(x H)^{- 1} = H^{- 1} x^{- 1} = H x^{- 1}$, so $x H \longmapsto H x^{- 1}$ is a bijection between the set of left cosets and the set of right cosets.

Exercise 1.1.6: In the last proof, if we said that $x H \longmapsto H x$ is a bijection between the set of left cosets and the set of right cosets, we would be lying. What is wrong with this statement?

We will denote by $[G : H]$ the number of left (or right) cosets of a subgroup $H$ of a group $G$. It can be finite even if $G$ and $H$ are infinite – for example, the group $2\mathbb{Z}$ of even integers is a subgroup of the additive group $\mathbb{Z}$ of integers, and its index is $2$.

Proposition 1.1.2: If $G$ and $H$ are finite groups, then $|H|$ divides $|G|$, and $[G : H] = \frac{|G|}{|H|} .$

Proof. (Click to Expand/Collapse)

We observe that $|x H| = |H|$ since $h \longmapsto x h$ is a bijection $H \longrightarrow x H$. Thus each coset has cardinality $|H|$ and $G$ is the disjoint union of the $[G : H]$ disjoint cosets. Therefore $|G| = [G : H] \cdot |H|$.

Exercise 1.1.7: Let $G$ be a group. Define a relation $\sim$ on $G$ by $x \sim y$ if and only if there exists $z \in G$ such that $y = z x z^{- 1}$. Prove that this relation, called conjugacy, is an equivalence relation. If $x \sim y$ we say that $x$ is conjugate to $y$, or (equivalently) that $x$ is a conjugate of $y$ or that $x$ and $y$ are conjugates (of each other).

If $G$ is a group and $K$ is a subgroup, then $K$ is called normal if whenever $x \in K$ and $y$ is conjugate to $x$, we have $y \in K$. If $K$ is normal, we write $K \trianglelefteq G$ (or $K \vartriangleleft G$ if $K$ is a proper subgroup).

Proposition 1.1.3: If $K$ is the kernel of a homomorphism $\phi : G \longrightarrow H$ between groups $G$ and $H$, then $K$ is a normal subgroup of $G$.

Proof. (Click to Expand/Collapse)

If $y$ is conjugate to $x$ we may write $y = g x g^{- 1}$ for some $g \in G$. If $x \in K$ then $\phi (x) = 1$. Now $\phi (y) = \phi (g) \cdot \phi (x) \cdot \phi (g^{- 1}) = \phi (g) \cdot 1 \cdot \phi (g)^{- 1} = 1$, so $y \in K$. This shows that $K$ is closed under conjugation, hence normal.

Exercise 1.1.8: Let $\phi : G \longrightarrow H$ be a group homomorphism. If $\phi$ is bijective, prove that the inverse map $\phi^{- 1} : H \longrightarrow G$ is also a homomorphism. If this is true, $\phi$ is called an isomorphism, and we say that $G$ is isomorphic to $H$ and we write $G \cong H$. Explain why isomorphism has the three properties of an equivalence relation: (reflexive) $G \cong G$, (symmetric) $G \cong H$ implies $H \cong G$ and (transitive) if $G \cong H$, $H \cong K$ then $G \cong K$. Thus the class of groups is divided into equivalence classes, which are called isomorphism classes.

If $G$ is a group and $g \in G$, let $\left\langle g \right\rangle =\{g^n |n \in \mathbb{Z}\}$. This is a group, obviously. It is called the cyclic subgroup of $G$ generated by $g$. The terms "cyclic'' and "generated by'' will be given more formal meanings later.

Exercise 1.1.9: Prove that if $\left\langle g \right\rangle$ is infinite then it is isomorphic to the additive group $\mathbb{Z}$.

Exercise 1.1.10: Prove that if $\left\langle g \right\rangle$ has finite order $d$, then $d$ is the smallest positive integer such that $g^d = 1$. Moreover, prove that $g^n = 1$ if and only if $d|n$. (Hint: Reverse the order of the two statements by first proving that there is an integer $d$ such that $g^n = 1$ if and only if $d|n$; then show $d = |G|$. Use the division algorithm for $\mathbb{Z}$, which is the statement that given any $n \in \mathbb{Z}$ we can write $n = q r + d$ with $0 \le d < r$.)

If $d$ is the integer in Exercise 1.1.10, then $d$ is called the order of $g$. Thus the order of the element $g$ equals the order of the cyclic subgroup it generates.

Exercise 1.1.11: Prove that if $p$ is a prime, and if $|G| = |H| = p$, then $G \cong H$.

Suppose that $G$ is a group and $H$ a subgroup. We will denote by $G / H$ the set of left cosets $x H$. It is in general not a group, but there is a special case where it is.

Proposition 1.1.4: Let $G$ be a group and $K$ a normal subgroup. If $x K$ and $y K$ are cosets, then so is $(x K) (y K)$; more precisely, $x K y K = x y K$.

Proof. (Click to Expand/Collapse)

We have $x K y K = x y (y^{- 1} K y) K.$ Since $K$ is normal, $y^{- 1} K y = K$ and so $(y^{- 1} K y) K = K$. Thus $x K y K = x y K$.

Because of Proposition 1.1.4, the set $G / K$ has a multiplicative structure, and is in fact a group. The identity element in $G / K$ is the coset $K = 1 \cdot K$. We call $G / K$ the quotient group.

In Proposition 1.1.3 we saw that the kernel of a homomorphism is normal; now let us observe that a normal subgroup is the kernel of a homomorphism.

Proposition 1.1.5: Let $K \trianglelefteq G$. Then $K$ is the kernel of the map $\pi : G \longrightarrow G / K$ defined by $\phi (g) = g K$, which is a surjective homomorphism.

We will sometimes call the map $\pi$ the projection onto the quotient group $G / K$.

Proof. (Click to Expand/Collapse)

The identity $x y K = x K y K$ in Proposition 1.1.4 can be written $\pi (x y) = \pi (x) \pi (y)$, so $\pi$ is a homomorphism. It is obviously surjective. If $x \in G$ then $x \in \ker (\pi)$ if and only if $x K = K$, that is, if $x \in K$, so the kernel is $K$.

Theorem 1.1.1: Let $\phi : G \longrightarrow H$ be any surjective group homomorphism, and let $K = \ker (\phi)$. Then $H \cong G / K$. More precisely, the rule $\lambda (g K) = \phi (g)$ gives a well-defined bijection $\lambda : G / K \longrightarrow H$, which is a group isomorphism.

Proof. (Click to Expand/Collapse)

We first prove that if $g, g' \in G$, then
 $\text{g K = g' K if and only if \phi (g) = \phi (g') .}$ (1.1.1)

Indeed, $g K = g' K$ if and only if $K = g^{- 1} g' K$, which is if and only if $g^{- 1} g' \in K$, which is if and only if $\phi (g^{- 1} g') = 1$, by definition of $K$ as the kernel of $\phi$. And since $\phi$ is a homomorphism, this is true if and only if $\phi (g)^{- 1} \phi (g') = 1$, that is, $\phi (g) = \phi (g')$. This proves (1.1.1).

Now (1.1.1) may be interpreted as the statement that $\lambda (g K) = \phi (g)$ is a well defined map $G / K \longrightarrow H$, and furthermore, it is injective. It is surjective since $\phi$ was assumed surjective. The fact that it is a group homomorphism is a consequence of the identity $x y K = x K y K$ in Proposition 1.1.4.

Proposition 1.1.6: Let $G$ be a cyclic group of order $n$. Then $G \cong \mathbb{Z}/ n\mathbb{Z}$ as an abelian group.

Note that $\mathbb{Z}$ is a group customarily written additively, and $n\mathbb{Z}$ is the subgroup of integer multiples of $n$. Since a cyclic group of order $n$ is determined by its order, it is good to have a consistent notation for it. The only completely standard notation for a cyclic group of order $n$ is $\mathbb{Z}/ n\mathbb{Z}$, but this has a connotation of being written additively, and sometimes the group would be written multiplicatively. We will denote a cyclic group of order $n$ by $Z_n$, though other authors might use the notation $C_n$.

Proof. (Click to Expand/Collapse)

Let $G = \left\langle g \right\rangle$, where $g$ has order $n$. There is a homomorphism $c : \mathbb{Z} \longrightarrow G$ given by $c (k) = g^k$. (Since we are writing $\mathbb{Z}$ additively and $G$ multiplicatively, "homomorphism'' means $c (n + m) = c (n) c (m)$.) The kernel of this homomorphism is the set of multiples of $n$ by Exercise 1.1.10, that is, $n\mathbb{Z}$. The statement now follows from Theorem 1.1.1.

Here is another application.

Proposition 1.1.7: Let $G$ be a group, $H$ and $K$ subgroups, and assume that $K \vartriangleleft G$. Then $H K$ is a group and $K$ is a normal subgroup of $H K$. Moreover $K \cap H$ is a normal subgroup of $H$, and
 $H K / K \cong H / (H \cap K) .$ (1.1.2)

Proof. (Click to Expand/Collapse)

We leave it to the reader to show that $H K$ is a group (Exercise 1.1.12). Obvious $K$ is normal in $H K$ since it is normal in the larger group $G$. Consider the homomorphism $\phi : H \longrightarrow H K / K$, which is the composite of the inclusion $H \longrightarrow H K$ with the projection $H K \longrightarrow H K / K$. This composite is surjective and the kernel is $H \cap K$ (Exercise 1.1.12). Now (1.1.2) follows from Theorem 1.1.1.

Exercise 1.1.12: Complete the proof of Proposition 1.1.7 by showng that $H K$ is a group, and that the homomorphism $H \longrightarrow H K / K$ is surjective, with kernel $H \cap K$.

Proposition 1.1.8: If $G$ is a group, and $S \subset G$ a subset of $G$, there is a unique smallest subgroup $H$ of $G$ containing $S$. Specifically, $H$ is a subgroup of $G$ satisfies the two conditions

• $H \supseteq S$;
• If $H'$ is any subgroup of $G$ containing $S$ then $H \subseteq H'$.
These conditions characterize $H$ uniquely.

The group $H$ is called the group generated by $S$, and is denoted $\left\langle S \right\rangle$. In particular if $S =\{x\}$ consists of a single element, we write $H = \left\langle x \right\rangle$ and call it a cyclic group with generator $x$. If $S =\{x_1, x_2, \cdots\}$ we may also write $\left\langle S \right\rangle = \left\langle x_1, x_2, \cdots \right\rangle$.

In the proof we will first prove the uniqueness assertion: that there cannot be more than one smallest group containing $S$. Then we will give two proofs of existence, one abstract, one constructive, illustrating two different ways of thinking about this. Either proof can be adapted to many other situations: if we are give a subset $S$ of a ring, field, vector space, etc. we may similarly consider the subring (or ideal), subfield, vector subspace, etc. generated by $S$.

Proof. (Click to Expand/Collapse)

First we prove uniqueness, then existence. Suppose $H_1$ and $H_2$ are both satisfy the two bulleted conditions. Then applying the two conditions first with $H = H_1$ and $H' = H_2$ gives $H_1 \subseteq H_2$, and symmetrically, $H_2 \subseteq H_1$. Hence $H_1 = H_2$.

We now give two constructions of a group $H$ satisfying the two properties. The first proof is to define $H = \bigcap_{\begin{array}{c} H' \supseteq S\\ \text{H' is a subgroup of G} \end{array}} H' .$ Note that the intersection is nonempty since $H' = G$ will be one of the subgroups in the intersection. Clearly $H$ is a subgroup of $G$, containing $S$, since all the $H'$ in the intersection contain it. If $H'$ is a subgroup of $G$ containing $S$, then $H'$ is among the groups in the intersection, so $H \subseteq H'$.

We give an alternative, more constructive proof based on different concepts. Define $H$ to be the set consisting of $1$ (just in case $S$ is empty) and all products $x_1 x_2 \cdots x_n$ such that either $x_i \in S$ or $x_i^{- 1} \in S$. It is clear that $H$ is closed under multiplication and the inverse map, so it is a subgroup. On the other hand if $H'$ is any group containing $S$ then since it is closed under multiplication and the inverse map, it obviously contains all the elements we have mandated to be in $H$, so $H \subseteq H'$.

We say that a subset $S$ of $G$ is a generating set if $G = \left\langle S \right\rangle$.

Exercise 1.1.13: Let $f, f' : G \longrightarrow H$ be a homomorphism, and let $S$ be a generating set for $G$. That is, assume that $f (s) = f' (s)$ for all $s \in S$. Prove that $f = f'$.