Define the set of symmetries of a rhombus as \[\mathscr{S}=\{e, R_{\text{ }O,\text{ }180^\circ}, r_{m_1}, r_{m_1}\}.\]
Claim \(\mathscr{S}\) is a group under composition.
We want to show that comosing two symmetries equals a symmetry.
Let \(\square 1234\) be a non-square rhombus, and suppose F and G are in \(\mathscr{S}\).
\[F\circ G(\square 1234)=F(G(\square 1234))=F(\square 1234)=\square 1234\]
For example:
Let \(F=r_{m_1}\) and \(G=r_{m_2}\). Then \(F\circ G(\square 1234)=R_{O,180^\circ}\).
img.
\(F\circ(G\circ H)=(F\circ G)\circ H\)
\[\begin{equation}\label{D14,1} \begin{split} F\circ(G\circ H) &= F\circ (G\circ H)(\square 1234)\\ &= F(G\circ H (\square 1234))\\ &= F(G(H(\square 1234)))\\ &= (F\circ G)\circ H(\square 1234) \end{split} \end{equation}\]
e
Example: \(R_{O,180^\circ}\circ e=R_{O,180^\circ}\)
Every symmetry of a rhombus undoes itself.
\(e\circ e=e\)
\(R_{O,180^\circ}\circ R_{O,180^\circ}=e\)
\(r_{m_1}\circ r_{m_1}=e\)
\(r_{m_2}\circ r_{m_2}=e\)