#
Section2.4Examples of groups/nongroups, Part I¶ permalink

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

#####
Example2.4.1

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

#####
Example2.4.2

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.

\(\langle \Q,+\rangle\)

\(\langle \Z,-\rangle\)

\(\langle \R,\cdot\rangle\)

\(\langle \C^*,\cdot\rangle\)

\(\langle \R,+\rangle\)

\(\langle \Z^+,+\rangle\)

\(\langle \Z^*,\cdot\rangle\)

\(\langle \M_n(\R),+\rangle\)

\(\langle \C,+\rangle\)

\(\langle \Z,\cdot\rangle\)

\(\langle \R^*,\cdot\rangle\)

\(\langle \M_n(\R),\cdot\rangle\)

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

#####
Definition2.4.3

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

\begin{equation*}
GL(n,\R)=\{M\in \M_n(\R):\det M \neq 0\}
\end{equation*}
(that is, let \(GL(n,\R)\) be the set of all invertible \(n \times n\) matrices over \(\R\)) , and let

\begin{equation*}
SL(n,\R)=\{M\in \M_n(\R):\det M =1\}\text{.}
\end{equation*}
These subsets of \(\M_n(\R)\) are called, respectively, the *general* and *special linear groups of degree \(n\) over \(\R\)*.

Note that these definitions imply that these subsets of \(\M_n(\R)\) are groups (under some operation). Sure enough, they are!

#####
Theorem2.4.4

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

##### Proof

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

We know that matrix multiplication is always associative,so \(\G_2\) holds. Next,

\begin{equation*}
I_n= \begin{bmatrix}1 \amp 0 \amp 0 \amp \cdots \amp 0 \\
0 \amp 1 \amp 0 \amp \cdots \amp 0 \\
0 \amp 0 \amp 1 \amp \cdots \amp 0 \\
\vdots \amp \vdots \amp \vdots \amp \ddots \amp \vdots \\
0 \amp 0 \amp 0 \amp \cdots \amp 1
\end{bmatrix},
\end{equation*}
the is in \(GL(n,\R)\)since \(\det I_n=1\neq 0\text{,}\) and it acts as an identity element for \(\langle GL(n,\R), \cdot\,\rangle\) since

\begin{equation*}
AI_n=I_nA = A
\end{equation*}
for all \(A\in GL(n,\R)\text{.}\)

Finally, let \(A\in GL(n,\R)\text{.}\) Since \(\det A\neq 0\text{,}\) \(A\) has (matrix multiplicative) inverse \(A^{-1}\) in \(\M_2(\R)\text{.}\) But we need to verify that \(A^{-1}\) is in \(G\text{.}\) This is in fact the case, however, since \(A^{-1}\) is invertible (it has inverse \(A\)), hence \(\det
A^{-1} \neq 0\text{.}\) Thus, \(A^{-1}\) is also in \(GL(n,\R)\text{.}\)

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

#####
Theorem2.4.5

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

##### Proof

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.

#####
Example2.4.7

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