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.