##### Theorem11.3.1Identities are Unique

The identity of a group is unique.

In this section, we will present some of the most basic theorems of group theory. Keep in mind that each of these theorems tells us something about every group. We will illustrate this point with concrete examples at the close of the section.

The identity of a group is unique.

One difficulty that students often encounter is how to get started in proving a theorem like this. The difficulty is certainly not in the theorem's complexity. It's too terse! Before actually starting the proof, we rephrase the theorem so that the implication it states is clear.

If \(G= [G; *]\) is a group and \(e\) is an identity of \(G\), then no other element of \(G\) is an identity of \(G\).

Next we justify the phrase “... *the* inverse of an element of a group.”

The inverse of any element of a group is unique.

The same problem is encountered here as in the previous theorem. We will leave it to the reader to rephrase this theorem. The proof is also left to the reader to write out in detail. Here is a hint: If \(b\) and \(c\) are both inverses of \(a\), then you can prove that \(b = c\). If you have difficulty with this proof, note that we have already proven it in a concrete setting in Chapter 5.

As mentioned above, the significance of Theorem 11.3.3 is that we can refer to *the* inverse of an element without ambiguity. The notation for the inverse of \(a\) is usually \(a^{-1}\) (note the exception below).

In any group, \(e^{-1}\) is the inverse of the identity \(e\), which always is \(e\).

\(\left(a^{-1}\right)^{-1}\) is the inverse of \(a^{-1}\) , which is always equal to \(a\) (see Theorem 11.3.5 below).

\((x*y*z)^{-1}\) is the inverse of \(x * y * z\).

In a concrete group with an operation that is based on addition, the inverse of \(a\) is usually written \(-a\). For example, the inverse of \(k - 3\) in the group \([\mathbb{Z}; +]\) is written \(-(k- 3)=3-k\). In the group of \(2 \times 2 \) matrices over the real numbers under matrix addition, the inverse of \(\left( \begin{array}{cc} 4 & 1 \\ 1 & -3 \\ \end{array} \right)\) is written \(-\left( \begin{array}{cc} 4 & 1 \\ 1 & -3 \\ \end{array} \right)\), which equals \(\left( \begin{array}{cc} -4 & -1 \\ -1 & 3 \\ \end{array} \right)\).

If a is an element of group \(G\), then \(\left(a^{-1}\right)^{-1}=a\).

Again, we rephrase the theorem to make it clear how to proceed.

If \(a\) has inverse \(b\) and \(b\) has inverse \(c\), then \(a = c\).

The next theorem gives us a formula for the inverse of \(a * b\). This formula should be familiar. In Chapter 5 we saw that if \(A\) and \(B\) are invertible matrices, then \((A B)^{-1}= B^{-1} A^{-1}\).

If \(a\) and \(b\) are elements of group \(G\), then \((a*b)^{-1}= b^{-1}*a^{-1}\).

If \(a\), \(b\), and \(c\) are elements of group \(G\), then \begin{equation*} \begin{array}{lc} \textrm{left cancellation:}& (a * b = a * c) \Rightarrow b = c\\ \textrm{right cancellation:}& (b * a = c * a) \Rightarrow b = c\\ \end{array} \end{equation*}

If \(G\) is a group and \(a, b \in G\), the equation \(a * x = b\) has a unique solution, \(x = a^{-1} * b\). In addition, the equation \(x * a = b\) has a unique solution, \(x = b * a^{-1}\).

Our proof of Theorem 11.3.9 was analogous to solving the concrete equation \(4x = 9\) in the following way: \[4 x=9=\left(4\cdot \frac{1}{4}\right)9=4\left(\frac{1}{4}9\right)\] Therefore, by cancelling 4, \[x = \frac{1}{4}\cdot 9 = \frac{9}{4}\]

If \(a\) is an element of a group \(G\), then we establish the notation that \[a * a = a^2\quad \quad a*a*a=a^3\quad \quad \textrm{etc.}\] In addition, we allow negative exponent and define, for example, \[a^{-2}= \left(a^2\right)^{-1}\] Although this should be clear, proving exponentiation properties requires a more precise recursive definition.

For \(n \geq 0\), define \(a^n\) recursively by \(a ^0 = e\) and if \(n > 0, a^n= a^{n-1} *a\). Also, if \(n >1\), \(a^{-n}= \left(a^n\right)^{-1}\).

In the group of positive real numbers with multiplication, \[5^3= 5^2\cdot 5 =\left(5^1\cdot 5\right)\cdot 5=\left(\left(5^0\cdot 5\right)\cdot 5\right)\cdot 5=((1\cdot 5)\cdot 5)\cdot 5= 5 \cdot 5\cdot 5=125\] and \[5^{-3}= (125)^{-1}= \frac{1}{125}\]

In a group with addition, we use a different form of notation, reflecting the fact that in addition repeated terms are multiples, not powers. For example, in \([\mathbb{Z}; +]\), \(a + a\) is written as \(2a\), \(a + a + a\) is written as \(3a\), etc. The inverse of a multiple of a such as \(- (a + a + a + a + a) = -(5a)\) is written as \((-5)a\).

Although we define, for example, \(a^5=a^4* a\), we need to be able to extract the single factor on the left. The following lemma justifies doing precisely that.

Let \(G\) be a group. If \(b\in G\) and \(n\geq 0\), then \(b^{n+1}=b* b^n\), and hence \(b* b^n= b^n*b\).

Based on the definitions for exponentiation above, there are several properties that can be proven. They are all identical to the exponentiation properties from elementary algebra.

If a is an element of a group \(G\), and \(m\) and \(n\) are integers,

\(a^{-n}= \left(a^{-1}\right)^n\) and hence \(\left(a^n\right)^{-1}= \left(a^{-1}\right)^n\)

\(a^{n+m}= a^n*a^m\)

\(\left(a^n\right)^m= a^{n m}\)

Our final theorem is the only one that contains a hypothesis about the group in question. The theorem only applies to finite groups.

If \(G\) is a finite group, \(\left| G\right| = n\), and \(a\) is an element of \(G\), then there exists a positive integer \(m\) such that \(a^m= e\) and \(m\leq n\).

Consider the concrete group \([\mathbb{Z}; +]\). All of the theorems that we have stated in this section except for the last one say something about \(\mathbb{Z}\). Among the facts that we conclude from the theorems about \(\mathbb{Z}\) are:

Since the inverse of 5 is \(-5\), the inverse of \(-5\) is 5.

The inverse of \(-6 + 71\) is \(-(71) + -(-6) = -71 + 6\).

The solution of \(12 + x = 22\) is \(x = -12 + 22\).

\(-4(6) + 2(6) = (-4 + 2)(6) = -2(6) = -(2)(6)\).

\(7(4(3)) = (7\cdot 4)(3) = 28(3)\) (twenty-eight 3s).

Let \([G; * ]\) be a group and \(a\) be an element of \(G\). Define \(f:G \to G\) by \(f(x) = a * x\).

Prove that \(f\) is a bijection.

On the basis of part a, describe a set of bijections on the set of integers.

Answer

Rephrase Theorem 11.3.3 and write out a clear proof.

Prove by induction on \(n\) that if \(a_1\), \(a_2\), $\ldots $, \(a_n\) are elements of a group \(G\), \(n\geq 2\), then \(\left(a_1*a_2*\cdots *a_n\right)^{-1}= a_n^{-1}*\cdots *a_2^{-1}*a_1^{-1}\). Interpret this result in terms of \([\mathbb{Z}; +]\) and \([\mathbb{R}^*;\cdot]\).

AnswerTrue or false? If \(a\), \(b\), \(c\) are elements of a group \(G\), and \(a * b = c * a\), then \(b = c\). Explain your answer.

Prove Theorem 11.3.14.

AnswerEach of the following facts can be derived by identifying a certain group and then applying one of the theorems of this section to it. For each fact, list the group and the theorem that are used.

\(\left(\frac{1}{3}\right)5\) is the only solution of \(3x = 5\).

\(-(-(-18)) = -18\).

If \(A, B, C\) are \(3\times 3\) matrices over the real numbers, with \(A + B = A + C\), then \(B = C\).

There is only one subset of the natural numbers for which \(K \oplus A = A\) for every \(A \subseteq N\).