2.4: Examples of Groups/Nongroups, Part I

Let's look at some examples of groups/nongroups.

Example \(\PageIndex{1}\)

We claim that \(\mathbb{Z}\) is a group under addition. Indeed, \(\langle\mathbb{Z},+\rangle\) is a binary structure and that addition is associative on the integers. The integer \(0\) acts as an identity element of \(\mathbb{Z}\) under addition (since \(a+0=0+a=a\) for each \(a\in \mathbb{Z}\)), and each element \(a\) in \(G\) has inverse \(-a\) since \(a+(-a)=-a+a=0\text{.}\)

Example \(\PageIndex{2}\)

For each following binary structure \(\langle G,*\rangle\text{,}\) determine whether or not \(G\) is a group. For those that are not groups, determine the first group axiom that fails, and provide a proof that it fails.

1. \(\langle \mathbb{Q},+\rangle\)
2. \(\langle \mathbb{Z},-\rangle\)
3. \(\langle \mathbb{R},\cdot\rangle\)
4. \(\langle \mathbb{C}^*,\cdot\rangle\)
5. \(\langle \mathbb{R},+\rangle\)
6. \(\langle \mathbb{Z}^+,+\rangle\)
7. \(\langle \mathbb{Z}^*,\cdot\rangle\)
8. \(\langle \mathbb{M}_n(\mathbb{R}),+\rangle\)
9. \(\langle \mathbb{C},+\rangle\)
10. \(\langle \mathbb{Z},\cdot\rangle\)
11. \(\langle \mathbb{R}^*,\cdot\rangle\)
12. \(\langle \mathbb{M}_n(\mathbb{R}),\cdot\rangle\)

If you have taken linear algebra, you have also probably seen a collection of matrices that is a group under matrix multiplication.

Definition: General and Special Linear Groups

Recall that given a square matrix \(A\text{,}\) the notation \(\det A\) denotes the determinant of \(A\text{.}\) Let

\begin{equation*} GL(n,\mathbb{R})=\{M\in \mathbb{M}_n(\mathbb{R}):\det M \neq 0\} \end{equation*}

(that is, let \(GL(n,\mathbb{R})\) be the set of all invertible \(n \times n\) matrices over \(\mathbb{R}\)) , and let

\begin{equation*} SL(n,\mathbb{R})=\{M\in \mathbb{M}_n(\mathbb{R}):\det M =1\}\text{.} \end{equation*}

These subsets of \(\mathbb{M}_n(\mathbb{R})\) are called, respectively, the general and special linear groups of degree \(n\) over \(\mathbb{R}\).

Note that these definitions imply that these subsets of \(\mathbb{M}_n(\mathbb{R})\) are groups (under some operation). Sure enough, they are!

Theorem \(\PageIndex{1}\)

\(GL(n,\mathbb{R})\) is a group under matrix multiplication.

Proof

Let \(A,B∈GL(n,\mathbb{R})\). Then \(\text{det}(AB)=(\text{det}A)(\text{det}B)≠0\) (since \(\text{det}A, \text{det}B≠0\)), so \(AB∈GL(n,\mathbb{R})\). Thus, \( \langle GL(n,\mathbb{R}),\cdot \rangle\) \(\langle GL(n,\mathbb{R}), \cdot \rangle\) is a binary structure.

We know that matrix multiplication is always associative, so \(\mathcal{G}_2\) holds. Next,

\(\begin{bmatrix} 1 & 0 & 0 & \cdots & 0\\ 0 & 1 & 0 & \cdots & 0\\ 0 & 0 & 1 & \cdots & 0\\ \vdots & \vdots & \vdots & \ddots & \vdots\\ 0 & 0 & 0 & \cdots & 1 \end{bmatrix}\)

the is in \(GL(n,\mathbb{R})\) since \(\text{det}I_n=1≠0\), and it acts as an identity element for \(\langle GL(n,\mathbb{R}),\cdot \rangle\) \(\langle GL(n,\mathbb{R}),\cdot \rangle\) since

\(AI_n=I_nA=A\)

for all \(A∈GL(n,\mathbb{R})\).

Finally, let \(A∈GL(n,\mathbb{R})\). Since \(\text{det}A≠0\), \(A\) has (matrix multiplicative) inverse \(A^{−1}\) in \(\mathbb{M}_2(\mathbb{R})\). But we need to verify that \(A^{−1}\) is in \(G\). This is in fact the case, however, since \(A^{−1}\) is invertible (it has inverse \(A\)), hence \(\text{det}A^{−1}≠0\). Thus, \(A^{−1}\) is also in \(GL(n,\mathbb{R})\).

So \(GL(n,\mathbb{R})\) is a group under multiplication.

Theorem \(\PageIndex{2}\)

\(SL(n,\mathbb{R})\text{,}\) is a group under matrix multiplication.

Proof

Let \(A,B∈SL(n,\mathbb{R})\). Then \(\text{det}(AB)=(\text{det}A)(\text{det}B)=1(1)=1\), so \(AB∈SL(n,\mathbb{R})\). Thus, \( \langle SL(n,\mathbb{R}),\cdot \rangle\) \(\langle SL(n,\mathbb{R}),\cdot \rangle\) is a binary structure.

The rest of the proof is left as an exercise for the reader.

Remark

Throughout this course, if we are discussing a set \(GL(n,\mathbb{R})\) or \(SL(n,\mathbb{R})\text{,}\) you should assume \(n\in \mathbb{Z}^+\text{,}\) unless otherwise noted.

Let \(A,B\in SL(n, \R)\text{.}\) Then \(\det(AB)=(\det A)(\det B) =1(1)=1\text{,}\) so \(AB\in SL(n,\R)\text{.}\) Thus, \(\langle SL(n,\R), \cdot\rangle\) is a binary structure.

The rest of the proof is left as an exercise for the reader.We end this section with a final example.

Example \(\PageIndex{3}\)

Define \(*\) on \(\mathbb{Q}^*\) by \(a*b=(ab)/2\) for all \(a,b\in \mathbb{Q}^*\text{.}\) Prove that \(\langle \mathbb{Q}^*,*\rangle\) is a group.

Solution

First, \(\langle \mathbb{Q}^∗ \rangle, \;\) is a binary structure, since \((ab)/2\) is rational and nonzero whenever \(a,b\) are rational and nonzero.

Next, we check that \(\mathbb{Q}^∗\) under \(∗\) satisfies the group axioms. Since multiplication is commutative on \(\mathbb{Q}, ∗\) is clearly commutative on \(\mathbb{Q}^∗\), and so our work to show \(\mathcal{G}_2\) and \(\mathcal{G}_3\) is marginally reduced.

First, associativity of \(∗\) on \(\mathbb{Q}^∗\) is inherited from associativity of multiplication on \(\mathbb{Q}^∗\).

Notice that the perhaps “obvious” choice, \(1\), is not an identity element for \(\mathbb{Q}^∗\) under \(∗\): for instance, \(1∗3= 3/2 ≠3\). Rather, \(e\) is such an identity element if and only if for all \(a∈\mathbb{Q}\) we have \(a=e∗a=(ea)/2\). We clearly have \(a=(2a)/2\) for all \(a∈\mathbb{Q}^∗\); so \(2\) acts as an identity element for \(\mathbb{Q}^∗\) under \(∗\).

Finally, let \(a∈\mathbb{Q}^∗\). Since \(a≠0\), it makes sense to divide by \(a\); then \(4/a∈\mathbb{Q}^∗\), with \(a∗(4/a)=(a(4/a))/2=2\).

Thus, \(\langle \mathbb{Q}^∗,∗\rangle\) is a group.

First, \(\langle\Q^*,\rangle\) is a binary structure, since \((ab)/2\) is rational and nonzero whenever \(a,b\) are rational and nonzero.

Next, we check that \(\Q^*\) under \(*\) satisfies the group axioms. Since multiplication is commutative on \(\Q\text{,}\) \(*\) is clearly commutative on \(\Q^*\text{,}\) and so our work to show \(\G_2\) and \(\G_3\) is marginally reduced.

First, associativity of \(*\) on \(\Q^*\) is inherited from associativity of multiplication on \(\Q^*\text{.}\)

Notice that the perhaps “obvious” choice, 1, is not an identity element for \(\Q^*\) under \(*\text{:}\) for instance, \(1*3=3/2 \neq 3\text{.}\) Rather, \(e\) is such an identity element if and only if for all \(a\in \Q\) we have \(a=e*a=(ea)/2\text{.}\) We clearly have \(a=(2a)/2\) for all \(a\in \Q^*\text{;}\) so \(2\) acts as an identity element for \(\Q^*\) under \(*\text{.}\)

Finally, let \(a\in \Q^*\text{.}\) Since \(a\neq 0\text{,}\) it makes sense to divide by \(a\text{;}\) then \(4/a\in \Q^*\text{,}\) with \(a*(4/a)=(a(4/a))/2=2\text{.}\)

Thus, \(\langle \Q^*,*\rangle\) is a group.