Problem source

Consider the following sets with the usual definition of multiplication appropriate to each. In each case you may assume that the multiplication
is associative. In each case state, giving adequate reasons, whether or not the set is a group. 
\begin{questionparts}
\item the complex numbers of unit modulus; 
\item the integers modulo 4; 
\item the matrices 
\[
\mathrm{M}(\theta)=\begin{pmatrix}\cos\theta & -\sin\theta\\
\sin\theta & \cos\theta
\end{pmatrix},
\]
where $0\leqslant\theta<2\pi$; 
\item the integers $1,3,5,7$ modulo 8; 
\item the $2\times2$ matrices all of whose entries are integers; 
\item the integers $1,2,3,4$ modulo 5. 
\end{questionparts}
In the case of each pair of groups above state, with reasons, whether
or not they are isomorphic.

Solution source

\begin{questionparts}
    \item $\{ z \in \mathbb{C} : |z| = 1\}$ is a group.
    \begin{enumerate}
        \item (Closure) $|z_1z_2| = |z_1||z_2| = 1$. Set is closed under multiplication
        \item (Associativity) Multiplication of complex numbers is associative
        \item (Identity) $|1| = 1$
        \item (Inverses) $| \frac{1}{z} | = \frac{1}{|z|} = \frac{1}{1} = 1$, the set contains inverses
        
    \end{enumerate}
    \item the integers $\pmod{4}$ are not a group under multiplication, $2$ has no inverse, since $0 \times k \equiv 0 \pmod{4}$
    \item The set of rotation matrices is a group:

    \begin{enumerate}
        \item (Closure) 
        \begin{align*}
             \begin{pmatrix}\cos\theta_1 & -\sin\theta_1\\
\sin\theta_1 & \cos\theta_1
\end{pmatrix} \begin{pmatrix}\cos\theta_2 & -\sin\theta_2\\
\sin\theta_2 & \cos\theta_2
\end{pmatrix} &= {\scriptscriptstyle\begin{pmatrix}\cos\theta_1 \cos \theta_2 - \sin \theta_1\sin \theta_2 & -\sin\theta_1\ \cos \theta_1 - \sin \theta_2\cos\theta_1\\
\sin\theta_1\ \cos \theta_1 + \sin \theta_2\cos\theta_1 & \cos\theta_1 \cos \theta_2 - \sin \theta_1\sin \theta_2
\end{pmatrix}} \\
&= \begin{pmatrix}\cos(\theta_1+\theta_2) & -\sin(\theta_1+\theta_2)\\
\sin(\theta_1+\theta_2) & \cos(\theta_1+\theta_2)
\end{pmatrix} 
        \end{align*}
        Since $\cos, \sin$ are periodic with period $2\pi$, we can find $\theta_3 = \theta_1 + \theta_2 + 2k\pi$ such that $0 \leq \theta_3 < 2 \pi$, so our set is closed
    \item (Associativity) Matrix multiplication is associative
    \item (Identity) Consider $\theta = 0$
    \item (Inverses) Consider $2\pi - \theta$
    \end{enumerate}

    \item $\{1, 3, 5, 7\} \pmod{8}$ is a group:

    \begin{tabular}{c|cccc}
         & 1 & 3 & 5 & 7  \\ \hline
         1 & 1 & 3 & 5 & 7  \\
         3 & 3 & 1 & 7 & 5  \\
         5 & 5 & 7 & 1 & 3  \\
         7 & 7 & 5 & 3 & 1  \\
    \end{tabular}
    \begin{enumerate}
        \item (Closure) See Cayley table
        \item (Associativity) Integer multiplication is associative
        \item (Identity) $1$
        \item (Inverses) $x \mapsto x$ (See Cayley table)
    \end{enumerate}
    \item $2\times2$ matrices are not a group, consider $\textbf{0} = \begin{pmatrix}
        0 & 0 \\ 0 & 0
    \end{pmatrix}$, then $\mathbf{0}\mathbf{M} = \mathbf{0}$ for all other matrices.
    \item \begin{tabular}{c|cccc}
         & 1 & 2 & 3 & 4  \\ \hline
         1 & 1 & 2 & 3 & 4  \\
         2 & 2 & 4 & 1 & 3  \\
         3 & 3 & 1 & 4 & 2  \\
         4 & 4 & 3 & 2 & 1  \\
    \end{tabular}
    \begin{enumerate}
        \item (Closure) See Cayley table
        \item (Associativity) Integer multiplication is associative
        \item (Identity) $1$
        \item (Inverses) $1 \mapsto 1, 2 \mapsto 3, 3 \mapsto 2, 4 \mapsto 4$ (See Cayley table)
    \end{enumerate}
\end{questionparts}



\begin{tabular}{c|c|c|c|c}
     & \textbf{(i)} & \textbf{(iii)} & \textbf{(iv)} & \textbf{(vi)}   \\ \hline
     \textbf{(i)} & $\checkmark$ & $\checkmark$   consider $z \mapsto \begin{pmatrix} \cos \arg (z) & - \sin \arg(z) \\ \sin \arg(z) & \cos \arg(z) \end{pmatrix}$ & not finite & not finite \\
     \textbf{(iii)} & & $\checkmark$ & not finite & not finite \\
     \textbf{(iv)} & & & $\checkmark$ & no element order $4$ \\
     \textbf{(vi)} & & & & $\checkmark$\\
\end{tabular}