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.