
MATH31052 Topology

0a Background material on group theory

The second part of the course on the fundamental group makes some use of group theory. For convenience I have summarized the basic ideas which I shall need. I hope that this summary will be sufficient for the few students taking the course who have not encountered group theory before.

0.24 Definition. A group is a non-empty set $$G$$ together with a function $$G\times G\to G$$ (a binary operation) written $$(g_1,g_2) \mapsto g_1\cdot g_2$$ which has the following properties.

•  It is associative which means that $$(g_1\cdot g_2)\cdot g_3 = g_1\cdot (g_2\cdot g_3)$$ for all $$g_1$$, $$g_2$$, $$g_3\in G$$.

•  It has an identity which means that there is a (unique) element $$e\in G$$ such that $$g\cdot e = g = e\cdot g$$ for all $$g\in G$$.

•  Inverses exist which means that for each element $$g\in G$$ there is a (unique) element $$g'$$ such that $$g\cdot g' = e = g'\cdot g$$.

A group is said to be abelian if the binary operation has the following additional property.

•  It is commutative which means that $$g_1\cdot g_2 = g_2\cdot g_1$$ for all

0.25 Examples. (a) The integers $$\Z$$, the rationals $$\Q$$ and the real numbers $$\R$$ each is an abelian group under addition $$(a,b)\mapsto a+b$$. The unit is $$0$$ and the inverse of an element $$a$$ is $$-a$$.

(b) The integers $$\Z$$, the rationals $$\Q$$ and the real numbers $$\R$$ are not groups under multiplication $$(a,b)\mapsto a\times b = ab$$. This binary operation is associative and commutative, 1 is the identity but not every element has an inverse: in $$\Z$$ only the elements $$\pm 1$$ have an inverse; in $$\Q$$ and $$\R$$ every non-zero element $$a$$ has an inverse $$a^{-1}$$ but 0 does not have an inverse.

(c) The non-zero rationals $$\Q ^*$$ and the non-zero real numbers $$\R ^*$$ are abelian groups under multiplication.

(d) The binary operation given by subtraction $$(a,b)\mapsto a-b$$ on the integers $$\Z$$ is not associative, doesn't have a unit (so that there is no idea of an inverse) and is not commutative.

(e) The set of $$n\times n$$ non-singular (invertible) matrices is a group, known as the general linear group under multiplication. The unit is the identity matrix and the inverse is given by the usual matrix inverse.

(f) The set of congruence classes of integers modulo $$n$$, $$\Z _n$$, is an abelian group under addition (see Eccles, Chapter 21 for this set and the definition of addition).

0.26 Remarks. (a) If a binary operation has a unit then it is necessarily unique for, given units $$e_1$$ and $$e_2$$, then $$e_1 = e_1\cdot e_2$$ (since $$e_2$$ is a unit) $$=e_2$$ (since $$e_1$$ is a unit).

(b) Given a set $$G$$ with binary operation with a unit, if an element $$g\in G$$ has in inverse then it is necessarily unique since, given inverses $$g'$$ and $$g$$, $$g' = g'\cdot e = g'\cdot (g\cdot g) = (g'\cdot g)\cdot g = e\cdot g = g$$.

(c) Associativity means that we can write compositions of more than elements without brackets: in a group the expression $$g_1\cdot g_2\cdot g_3$$ is unambiguous since $$(g_1\cdot g_2)\cdot g_3 = g_1\cdot (g_2\cdot g_3)$$.

(d) Notations like $$(g_1,g_2)\mapsto g_1\cdot g_2$$ or $$g_1\ast g_2$$ are usually used initially when discussing a general binary operation. In practice when discussing groups the binary operation is often called the product or multiplication (represented by $$(g_1,G_2)\mapsto g_1 \times g_2$$ or just $$g_1g_2$$ with the identity either denoted either $$e$$ or $$1$$ and the inverse of $$g$$ denoted $$g^{-1}$$) or if the group is abelian it is called the sum or addition (represented by $$(g_1,g_2)\mapsto g_1+g_2$$ with identity denoted either $$e$$ or $$0$$ and the inverse of $$g$$ denoted $$-g$$). The notes for the course use the product notation.

0.27 Definition. A function $$f\colon G_1 \to G_2$$ between two groups is called a homomorphism if $$f(g_1 g_2) = f(g_1) f(g_2)$$ for each $$g_1$$, $$g_2\in G$$.

An injection $$f\colon G_1\to G_2$$ which is a homomorphism is called a monomorphism.

A surjection $$f\colon G_1\to G_2$$ which is a homomorphism is called an epimorphism.

A bijection $$f\colon G_1\to G_2$$ which is a homomorphism is called an isomorphism. If there is an isomorphism between two groups $$G$$ and $$G_2$$ then we say that they are isomorphic and write $$G_1\cong G_2$$.

0.28 Proposition. (a) If $$f\colon G_1\to G_2$$ is a homomorphism of groups then, if $$e_1$$ is the identity element of $$G_1$$, $$f(e_1)$$ is the identity element of $$G_2$$ and, for $$g\in G_1$$, $$f(g^{-1})$$ is the inverse of $$f(g)$$ in $$G_2$$.

(b) Given homomorphisms of groups $$f_1\colon G_1\to G_2$$ and $$f_2\colon G_2\to G_3$$, then the composition $$f_2\circ f_1\colon G_1\to G_3$$ is a homomorphism.

(c) If $$f\colon G_1\to G_2$$ is an isomorphism of groups then the inverse map $$f^{-1}\colon G_2\to G_1$$ is a homomorphism (and so an isomorphism).

Proof. (a) Write $$e_i$$ for the unit of $$G_i$$ ($$i=1,2$$). Since $$e_1 e_1 = e_1$$ it follows that $$f(e_1) f(e_1) = f(e_1)$$. Hence $$f(e_1) = f(e_1) e_2 = f(e_1) f(e_1) f(e_1)^{-1} = f(e_1) f(e_1)^{-1} = e_2$$.
For $$g\in G$$, $$f(g) f(g^{-1}) = f(g g^{-1}) = f(e_1) = e_2$$ and, similarly, $$f(g^{-1}) f(g) = e_2$$ so that $$f(g^{-1}) = f(g)^{-1}$$.

(b) For $$g_1$$, $$g_2\in G$$, $$(f_2\circ f_1)(g_1 g_2) = f_2\bigl (f_1(g_1 g_2)\bigr ) = f_2\bigl (f_1(g_1) f_1(g_2)\bigr )$$ (since $$f_1$$ is a homomorphism) $$= f_2\bigl (f_1(g_1)\bigr ) f_2\bigl (f_1(g_2)\bigr )$$ (since $$f_2$$ is a homomorphism) $$= (f_2\circ f_1)(g_1)(f_2\circ f_1)(g_2)$$. Hence $$f_2\circ f_1$$ is a homomorphism.

(c) Given $$h_1$$, $$h_2\in G_2$$, since $$f$$ is a bijection $$h_i=f(g_i)$$ for a unique $$g_i\in G_1$$ ($$i=1,2$$). Then $$f^{-1}(h_i)=g_i$$. Since $$f(g_1 g_2) = f(g_1) f(g_2) = h_1 h_2$$ and $$f$$ is a bijection, $$f^{-1}(h_1) f^{-1}(h_2)=g_1 g_2 = f^{-1}(h_1 h_2)$$ as required to prove that $$f^{-1}$$ is a homomorphism. Since it is also a bijection (the inverse of a bijection) it is an isomorphism.  $$\Box$$

0.29 Definition. Given a group $$G$$ a subset $$H\subset G$$ is a subgroup of $$G$$ if it is a group under the restriction of the binary operation on $$G$$ to $$H$$.

0.30 Examples. (a) The additive group of the integers $$\Z$$ is a subgroup of the additive group of the rationals $$\Q$$.

(b) The even integers form a subgroup of the additive group of the integers.

(c) The singleton subset of $$G$$ consisting of the identity $$\{e\}$$ is a subgroup of $$G$$ called the trivial subgroup. This subgroup is denoted by $$I$$.

0.31 Definition. Given a homomorphism of groups $$f\colon G_1\to G_2$$, the kernel of $$f$$ is defined by $$\ker (f)=\{\,g\in G_1\mid f(g) = e\,\}$$ where $$e$$ is the identity element of $$G_2$$.

0.32 Proposition. (a) The kernel $$\ker (f)$$ of a group homomorphism $$f\colon G_1\to G_2$$ is a subgroup of $$G_1$$.

(b) A group homomorphism $$f\colon G_1 \to G_2$$ is a monomorphism if and only if $$\ker (f) = I$$, the trivial subgroup of $$G_1$$.

Proof. (a) This is a simple check. [In what follows the identity elements of $$G_1$$ and $$G_2$$ are each denoted $$e$$ as is usual. It is clear from the context which identity element is meant.] Given $$g_1$$, $$g_2\in \ker (f)$$, $$f(g_1 g_2) = f(g_1) f(g_2) = e e = e$$ and so $$g_1 g_2\in \ker (f)$$. Since $$f(e)=e$$, it follows that $$e\in \ker (f)$$ and, for $$g\in \ker (f)$$, $$f(g^{-1}) = f(g)^{-1} = e^{-1} = e$$ and so the inverse $$g^{-1}\in \ker (f)$$. Hence $$\ker (f)$$ is a subgroup of $$G_1$$.

(b) Since $$f(e)=e$$, if $$f$$ is a monomorphism and so an injection then $$\ker (f) = \{e\} = I$$. Conversely, if $$\ker (f) = I$$, then given $$g_1$$, $$g_2\in G_1$$ such that $$f(g_1)=f(g_2)$$ it follows that $$f(g_1 g_2^{-1}) = f(g_1) f(g_2^{-1}) = f(g_2) f(g_2)^{-1} = e$$ and so $$g_1 g_2^{-1}\in \ker (f)$$ which means that $$g_1 g_2^{-1} = e$$ giving $$g_1=g_2$$ proving that $$f$$ is an injection and so a monomorphism.  $$\Box$$

0.33 Proposition. Given groups $$G$$ and $$H$$, a group structure may be put onto the cartesian product $$G\times H$$ by $$(g_1,h_1) (g_2,h_2) = (g_1 g_2, h_1 h_2)$$.

Proof. It is straightforward to check the conditions. The identity is given by $$(e,e)$$ and the inverse of $$(g,h)$$ is given by $$(g^{-1},h^{-1})$$.  $$\Box$$

#### References

•  P.J. Eccles, An Introduction to Mathematical Reasoning: numbers, sets and functions, Cambridge University Press, 1997, chapter 21.

pje\teaching\MATH31051\notes0a.tex December 21, 2020