We will use the notation \(\ast \colon S\times S\to S\) to denote a binary operation on a set \(S\) that sends the pair \((x,y)\) to \(x\ast y\text{.}\) Recall that a binary operation \(\ast\) is associative means that \(x\ast(y\ast z)= (x\ast y)\ast z\) for all \(x,y,z\in
S\text{.}\)
For every \(g\in G\text{,}\) there exists an element \(h\in G\text{,}\) called an inverse element for \(g\text{,}\) such that \(g\ast h=h \ast g=e\text{.}\)
Proposition2.2.2.Immediate consequences of the definition of group.
Let \(G\) be a group. The element \(e\) in the second property of Definition 2.2.1 is unique. Given \(g\in G\text{,}\) the element \(h\) in the third property of Definition 2.2.1 is unique.
Let \(G\) be a group. By Proposition 2.2.2, we may speak of an identity element as the identity element for \(G\text{.}\) Given \(g\in G\text{,}\) we may refer to an inverse element for \(g\) as the inverse of \(g\text{,}\) and we write \(g^{-1}\) to denote this element. In practice, we often omit the operator \(\ast\text{,}\) and simply write \(gh\) to denote \(g\ast h\text{.}\) We adopt the convention that \(g^0\) is the identity element. For \(k\geq
1\text{,}\) we write \(g^k\) to denote \(\underbrace{g\ast g\ast \cdots \ast g}_{k \text{
factors}}\) and we write \(g^{-k}\) to denote \(\left(g^{k}\right)^{-1}\text{.}\) This set of notational conventions is called multiplicative notation.
In general, group operations are not commutative (see Exercise 2.1.9.3). A group with a commutative operation is called Abelian. For some Abelian groups, such as the group of integers, the group operation is called addition, and we write \(a+b\) instead of using the multiplicative notation \(a\ast b\text{.}\) We write \(0\) to denote the identity element, we write \(-a\) to denote the inverse of \(a\text{,}\) and we write \(ka\) to denote \(\underbrace{a+ a+ \cdots +a}_{k
\text{ summands}}\) for positive integers \(k\text{.}\) This set of notational conventions is called additive notation.
The number of elements in a finite group is called the order of the group. A group with infinitely many elements is said to be of infinite order. We write \(|G|\) to denote the order of the group \(G\text{.}\)
A group with a single element (which is necessarily the identity element) is called a trivial group. In multiplicative notation, one might write \(\{1\}\text{,}\) and in additive notation, one might write \(\{0\}\text{,}\) to denote a trivial group.
Let \(G\) be a group. Suppose that \(e,e'\) both satisfy the second property of the Definition 2.2.1, that is, suppose \(e\ast x=x\ast e = e'\ast x=x\ast e'=x\) for all \(x\in G\text{.}\) Show that \(e=e'\text{.}\)
Let \(G\) be a group with identity element \(e\text{.}\) Let \(g\in G\) and suppose that \(g\ast h = h\ast g = g\ast h' = h'\ast g =
e\text{.}\) Show that \(h=h'\text{.}\)
Suppose that \(gx=hx\) for some elements \(g,h,x\) in a group \(G\text{.}\) Show that \(g=h\text{.}\) [Note that the same proof, mutatis mutandis, shows that if \(xg=xh\text{,}\) then \(g=h\text{.}\)]
Given two groups \(G,H\) with group operations \(\ast_G, \ast_H\text{,}\) the Cartesian product \(G\times H\) is a group with the operation \(\ast_{G\times H}\) given by
For part (a), let \(n\) be the least positive integer such that \(g^n=e\) (explain why \(n\) exists!). Given an arbitrary element \(h\in G\text{,}\) write \(h=g^k\) for some \(k\text{,}\) then use the Division Algorithm.
Let \(k,n\) be positive integers with \(k\leq n\text{,}\) and let \(A=\{a_1,a_2,\ldots,a_k\}\) be a set of \(k\) distinct elements in \(\{1,2,\ldots,n\}\text{.}\) We write \((a_1a_2\cdots a_k)\) to denote the permutation \(\sigma\) in \(S_n\) (see Definition 2.1.1) given by the assignments
and \(\sigma(j)=j\) for \(j\not\in A\text{.}\) If \(k=1\text{,}\) the permutation \((a_1)\) is the identity permutation. A permutation of the form \((a_1a_2\cdots a_k)\) is called a \(k\)-cycle. For example, the element \(\sigma=[1,4,2,3,5]=(2\;4\;3)\) is a 3-cycle in \(S_5\) because \(\sigma\) acts on the set \(A=\{2,3,4\}\) by
and \(\sigma\) acts on \(\{1,5\}\) as the identity. Note that cycle notation is not unique. For example, we have \((2\;4\;3)=(4\;3\;2)=(3\;2\;4)\text{.}\) A permutation is called cyclic if it is a \(k\)-cycle for some \(k\text{.}\) A 2-cycle is called a transposition.
Find all of the cyclic permutations in \(S_3\text{.}\) Find their inverses.
Cycles \((a_1a_2\cdots a_k)\) and \((b_1b_2\cdots b_\ell)\) are called disjoint if the sets \(\{a_1,a_2,\ldots,a_k\}\) and \(\{b_1,b_2,\ldots,b_\ell\}\) are disjoint, that is, if \(a_i\neq b_j\) for all \(i,j\text{.}\) Show that every permutation in \(S_n\) is either a cycle or a product of disjoint cycles.
Show that factoring a permutation into a product of transpositions is not unique by writing the identity permutation in \(S_3\) as a product of transpositions in two different ways.
The Cayley table for a finite group \(G\) is a two-dimensional array with rows and columns labeled by the elements of the group, and with entry \(gh\) in position with row label \(g\) and column label \(h\text{.}\) Partial Cayley tables for \(S_3\) (Figure 2.2.7) and \(D_4\) (Figure 2.2.8) are given below.
Prove that the Cayley table for any group is a Latin square. This means that every element of the group appears exactly once in each row and in each column.
