Elementary Properties of Groups

Property 1   Generalized associative law: We shall not give a careful formulation of this property nor shall we prove it (the interested reader can consult [Sc], p. 3, 4). This property essentially means that parentheses can be inserted or deleted at will (just as long as the order is not changed) without affecting the value of a product involving any number of group elements; e.g., if $a,b,c,d \in G$, $G$ a group, then $aa(bcdbb) = (aa)b(cd)bb = ((aa)b)(cd)(bb)$, etc.

Property 2   Uniqueness of the identity element: We claim that the element $e$ of condition 2 of Definition 2.1.2 is unique.

Proof: For suppose that $f$ is also an identity of $G$; i.e., $af = fa = a$ for all $a \in G$. Then $ef = e$, but on the other hand $ef = f$, since $e$ is an identity element. Consequently, $e = f$. $\Box$

Property 3   Uniqueness of the inverse element: We claim that for each element $a$ in $G$ the element $a^{- 1}$ of condition 3 of Definition 2.1.2 is unique.

Proof: Namely, suppose $ab = ba = e$ and $ac = ca = e$. Then $b = be = b(ac) = (ba)c = ec = c$. For each $a \in G$ we call this unique element $a^{- 1}\in G$. $\Box$

Property 4   If $a$ and $b$ are elements of a group $G$, then there exist unique elements $x$ and $y$ of $G$ such that $xa = b$ and $ay =b$.

Proof: If $x_0a = b$, then multiplying both sides on the right by $a^{- 1}$, yields $(x_0a)a^{- 1} = ba^{- 1}$ or that $x_0 = ba^{- 1}$. Conversely, if $x = ba^{- 1}$, then $xa = (ba^{- 1})a = b(a^{- 1}a) = be = b$. Hence, the equation $xa = b$ has the unique solution $x = ba^{- 1}$. Similarly, one shows that the equation $ay =b$ has the unique solution $y = a^{- 1}b$. $\Box$

Property 5   Alternate group definition (A): If $G$ is a semi-group in which the equations $xa = b$ and $ay =b$ are solvable for arbitrary $a,b \in G$, then $G$ is a group.

Proof: Let $e$ be a solution of the equation $xa = a$. Thus $ea = a$. Moreover, for any $b \in G$, there exists an element $y\in G$ such that $ay =b$. Now $eb = e(ay) = (ea)y = ay = b$. This shows that there exists an element $e \in G$ such that $eb = b$ for any $b \in G$. Now, analogously, consider the equation $ay = a$ and let $f$ be a solution so that $af = a$. Then for any $b \in G$ there exists an $x\in G$ such that $xa = b$; thus $bf = (xa)f = x(af) = xa = b$, and we have shown that there exists an element $f \in G$ such that $bf = b$ for any $b \in G$. Thus, we have

$$ef = f\ \ \ \ {\rm and}\ \ \ \ ef = e,$$

so $e = f$, and consequently $G$ contains an identity element. Now it follows from the hypothesis, that there exist $x_0,y_0 \in G$ such that $x_0a = e$ and $ay_0 = e$. Hence, $y_0 = ey_0 = (x_0a)y_0 = x_0(ay_0) = x_0e = x_0$, so $x_0 = y_0 = a^{- 1}$. This shows that the given statement is sufficient for $G$ being a group. The fact that it is also necessary is a consequence of Property 4 of our elementary group properties. $\Box$

Property 6   Cancellation laws: If $G$ is any group, then

(a)

(Left Cancellation) $wx = wy$ implies $x = y$ for $w,x,y \in G$.

(b)

(Right Cancellation) $xz = yz$ implies $x = y$ for $x,y,z \in G$.

Proof: Multiply both sides of (a) by $w^{-1}$ on the left; (b) is done similarly. $\Box$

We note that as a consequence of the cancellation laws, if we write the Cayley table for $G$ there will be no duplications in any row or column. As a matter of fact, this property of groups is quite useful to keep in mind when constructing the table in the first place. (See also Exercise 8 for this chapter.)

Property 7   Alternate finite group definition: A finite semi-group (i.e., a semi-group with a finite number of elements) in which the cancellation laws hold is a group.

Proof: Clearly this property is necessary for being a group from property 6. Now suppose that $A = \{a_1,a_2, ..., a_n\}$ is a finite semi-group satisfying the cancellation laws (see Property 6). Let $a$ be an arbitrary element of $A$. The $n$ elements $aa_1, ..., aa_n$ are then all distinct by the left cancellation law. Hence, if $b$ is an arbitrary element of $A$, then there exists an $a_i$ such that $aa_i = b$, i.e., the equation $ay =b$ is solvable. Similarly, the equation $xa = b$ is solvable in $A$, and therefore, by Property 5, $A$ is a group. $\Box$

Property 8   Alternate group definition (B): If $G$ is a semi-group which has at least one element $e \in G$ such that $ae = a$ for all $a \in G$ (such an element is called a right identity), and, if among all such elements $e$, there is an element $f$ such that to each $a \in G$ there exists an element $a^* \in G$ such that $aa^* = f$ (such an element is called a right inverse), then $G$ is a group.

Proof: If $G$ is a group, it is clear that these conditions are satisfied. Now suppose $G$ is a semi-group satisfying our conditions. Let $aa^* = f$. Then $faa^* = ff = f = aa^*$. Now there exists an $a^{**}\in G$ such that $a^*a^{**} = f$. Thus, $faa^*a^{**} = aa^*a^{**}$, or $faf = af$, so $fa = a$ since $f$ is a right identity. Thus $f$ is an identity, i.e., $af = fa = a$ for all $a \in G$. Thus if $G$ is a group then $f = e$ is the unique identity element. To prove that $G$ is a group, let $aa^* = f$. Then

$$\begin{array}{c} a^*a a^*=a^*,\\ a^* a a^* a^{**}= a^* a^{**},\\ a^* a f=f,\\ a^* a =f. \end{array}$$

Hence $f = e$ is the unique identity and $a^* = a^{- 1}$ is the unique inverse of $a$. $\Box$

Property 9   Laws of exponents: By (1), we know that we can unambiguously write $a_1a_2 ... a_n$ where all the $a_i \in G$, $G$ a group. If all the $a_i = g$, one writes this expression as $g^n$ and speaks of the $n^{th}$ power of $g$. (Note: $g$ may not be a number. So even though $g\in G$ and $g^n \in G$ it may be that $n \notin G$, e.g., if $G = GL(n,\mathbb {R})$.) Negative powers of $g$ can be defined as follows:

$$g^{-n} = (g^n)^{- 1} = (g^{- 1})^n.$$

(Note: If we just defined $g^{-n} = (g^{- 1})^n$, then it can be proven by induction on $n$ that $g^{- n} = (g^n)^{- 1}$ for all $n \in \mathbb {N}$.) Finally, one defines $g^0 = e$. It is then not hard to show that for m,n arbitrary integers, the following laws of exponents hold in $G$:

$$g^mg^n=g^ng^m=g^{m+n},$$ (2.2)

$$(g^m)^n=g^{mn}.$$ (2.3)

In the case of an abelian group $G$ written with the binary operation $+$, for $n\in\mathbb {Z}_+$ and $a \in G$, one writes $na$ instead of $a^n$, $na=a+...+a$ ($n$ times), and $(- n)a = - (na) = n(- a)$. The laws corresponding to (2.2) and (2.3) become for abelian groups

$$ma+na=na+ma=(m+n)a,$$

$$n(ma)=(mn)a,$$

where $m,n \in\mathbb {Z}$.

Definition 2.2.1   Consider now an element $g\in G$, a group. If all the powers, $g^n$ ($n=0,1,2,...$), of the element are distinct, then $g$ is called an element of infinite order in $G$.

Let us suppose that this is not the case. So there exist $m,n$, where $m \not= n$, say $m > n$, such that $g^m = g^n$. Then

$$g^{m- n} = e,$$

where $m - n > 0$. In other words, if $g$ is not an element of infinite order, then there exist positive integers $k$ such that $g^k = e$.

Definition 2.2.2   Let $G$ be a group and $a \in G$. Let $n$ be the smallest positive integer, if it exists, such that $a^n =e$ then $n$ is called the order of $a$ and we shall write $o(a) = n$. One also says that $a$ is of finite order with order $n$.

If $o(a) = n$, then all the elements

$$e,a,a^2, ..., a^{n- 1}$$ (2.4)

are distinct. For just as above, if any were equal we would get $a^t = e$ for $t < n$ in contradiction to the definition of $n$. Moreover, we also contend that any power $a^k$ is equal to one of the elements in (2.4). For the Division Algorithm gives that $k = nq + r$, $0 \leq r < n$. Then

$$a^k = (a^n)^q a^r = a^r,$$

by the laws of exponents. In addition, we see from this same relationship that if $o(a) = n$, and $a^k = e$, then $n\vert k$. Indeed, $r < n$, $a^r = e$, and $n$ is the smallest positive integer such that $a^n =e$, we must have $r = 0$. Thus $n\vert k$.

In summary, we have our last elementary property.

Property 10   If $G$ is a group, $a \in G$, and $o(a) = n < \infty$, then $e,a, ..., a^{n- 1}$ are distinct, any power of $a$ is equal to one of these, and finally $a^k = e$ if and only if $n\vert k$.

We have seen (Example 2.1.12) that there exist infinite groups all of whose elements have finite orders; such groups are called periodic. In any group, $G$, the identity $e$, of course, has finite order $1$. If this is the only element of $G$ with finite order, then $G$ is called torsion free.

We conclude this chapter with an important definition, viz., the notion of a subgroup of a group $G$. We shall make use of this concept throughout the text.

Definition 2.2.3   A nonempty subset H of a group G is called a subgroup of G if

(a)

$a,b \in H$ implies that $ab \in H$,

(b)

$e \in H$ (where $e$ is the identity of $G$),

(c)

$a \in H$ implies that $a^{- 1} \in H$.

Notation: We write

$$H \leq G,$$

when $H$ is a subgroup of $G$.

It is clear that a subgroup $H$ of a group $G$ is itself a group with respect to the same binary operation given on $G$. The definition can be given in a more succinct fashion, but we refer the reader to the exercises for this and related matters. We now list a few examples of subgroups of some of the groups given earlier in this chapter. Many more examples of subgroups will be encountered in the course of our investigations.

Example 2.2.4   Let $G = \mathbb {Q}$ (Example 2.1.9) with binary operation to be addition of rationals and let $H = \mathbb {Z}$. Clearly $H \leq G$. $\Box$

Example 2.2.5   Let $G = GL(n,\mathbb {R})$ be the group of Example 2.1.13 and take $H = \{A \in G \ \vert\ \det (A) = 1\}$. Then $H \leq G$. ($H$ is called the special linear group, denoted $SL(n,\mathbb {R})$.) $\Box$

Example 2.2.6   Take $G =\mathbb {C}^n$ (Example 2.1.14) and $H$ to be those $n$-tuples for which the first entry is $0$. Then $H \leq G$. $\Box$

Example 2.2.7   Let $G = \{1, -1, i, -i\}$, i.e., the $4^{th}$ roots of unity (see Example 2.1.11) and take $H = \{1, -1\}$. Then $H \leq G$. $\Box$