4.3 More properties of groups

Definition 4.7 Let \((G,*)\) be a group and \(g,h \in G\). We say that \(g,h\) commute if \(g*h=h*g\).

Definition 4.8 A group \((G,*)\) is called abelian or commutative if every two elements of G commute.

Example 4.3

\((\ZZ,+)\) is abelian, as are \((\CC^\times), \times\) and \((\ZZ_n, +)\), and \((\zz_p^\times, \times)\), and the trivial group.
\((S_1,\circ)\) and \((S_2,\circ)\) are abelian, but \((S_n,\circ)\) is not abelian if \(n\geq 3\) as \((1,2)(2,3) \neq (2,3)(1,2)\).
\(GL(n,\CC)\) is not abelian if \(n>1\).

You should verify the last statement by finding, for each \(n>1\), two invertible complex \(n\times n\) matrices that don’t commute.

Definition 4.9 Let \((G,*)\) be a group and let \(g \in G\).

The smallest positive integer n such that \(g^n = e\) is called the order of g.
If no such n exists, we say g has infinite order.

Don’t confuse the order of an element with the order of a group.

Lemma 4.2 Let \((G,*)\) be a finite group. Then every element of G has finite order.

Proof. Let \(g \in G\), and consider the elements \[\begin{equation*} g, g^2, g^3, \ldots \end{equation*}\] of G. Since G is finite and this list is infinitely long, the elements of the list can’t all be different: we must have \(g^a = g^b\) for some \(a<b\). Then \(g^{b-a}=e\), that is, some positive power of g equals the identity element and so g has finite order.

Lemma 4.3 Suppose \((G,*)\) is a group and \(g \in G\) has order n. Then the elements \(e=g^0, g, g^2,\ldots,g^{n-1}\) are all different.

Proof. Suppose \(g^i=g^j\) for some \(0 \leq i < j \leq n-1\). Then we can write \(j=i+k\) for some \(0<k<n\). So \(g^i = g^{i+k}=g^ig^k\). Multiplying both sides by \((g^i)^{-1}=g^{-i}\) we get \(e = g^k\). But this contradicts n being the smallest positive power of g which equals the identity.

Definition 4.10

A group \((G,*)\) is called cyclic if there is a \(g \in G\) such that \(G = \{ g^n : n \in \ZZ\}\).
An element \(g\in G\) such that \(G = \{g^n :n \in \ZZ\}\) is called a generator of \((G,*)\).

So G is cyclic if it has an element g such that any element of G is equal to a power of g.

Lemma 4.4 Cyclic groups are abelian.

Proof. Let \((G,*)\) be a cyclic group and g be a generator. Any element of G is equal to some power of g, but if \(i,j \in \ZZ\) then \[ g^i g^j = g^{i+j} = g^{j+i} = g^j g^i \] so powers of g commute with each other. Thus any two elements of G commute, and G is abelian.

Example 4.4

\((\ZZ, +)\) is cyclic, and 1 and -1 are both generators. 1 is a generator because any nonzero element of \(\ZZ\) can be obtained by adding some number of 1s together or some number of -1s together. -1 is a generator for the same reason.
\((\ZZ_n, +)\) is cyclic, and 1 is a generator. For the elements of \(\ZZ_n\) are \(0, 1, 2=1+1, 3=1+1+1,\ldots\).
\(GL(2,\CC)\) can’t be cyclic, because it is not even abelian.
\((\{1,-1,i,-i\},\times)\) is cyclic, and i is a generator.
\((\{ [1]_{15},[2]_{15},[4]_{15},[7]_{15},[8]_{15},[11]_{15},[13]_{15},[14]_{15}\}, \times)\) is not cyclic, as you can see by checking each group element in turn to see that it is not a generator.
\((\mathbb{Q},+)\) is abelian, isn’t cyclic. Why not?
Let \(C_n = \{ e^{2\pi i k /n} : k \in \ZZ\}\), a subset of the complex numbers. This is a group under multiplication: certainly multiplication is a binary operation on this set, for \[\begin{equation*} e^{2\pi i k/n}e^{2\pi i l/n}=e^{2\pi i(k+l)/n} \end{equation*}\] which is an element of \(C_n\). You can check the other group axioms. \(C_n\) is a cyclic group, because every element is a power of \(\zeta = e^{2\pi i/n}\), and \(\zeta\) has order n so \(|C_n|= n\). Any generator of \(C_n\) is called a primitive nth root of unity.