## 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*n*th root of unity.