For any positive integer \(n\), let \(G_n\) be the group generated by \(a\) and \(b\) subject to the following three relations: \[\begin{align*} a^2=1, \quad b^2=1, \quad \text{and} \quad (ab)^n=1. .\end{align*}\]
Find the order of the group \(G_n\).
Classify all irreducible complex representations of \(G_4\) up to isomorphism.
Solution
1. Put \(r=ab\). Then \(b=ar\), and\[ a r a=r^{-1}. \]
Since \(a^2=1\), \(b=ar\). Also \[ ara=a(ab)a=ba=(ab)^{-1}=r^{-1}. \] Thus \(G_n\) is generated by \(a,r\) with relations \[ a^2=1,\qquad r^n=1,\qquad ara=r^{-1}. \]
2. Every element of \(G_n\) has the form \(r^j\) or \(ar^j\), with \(0\le j<n\). Hence \(|G_n|\le 2n\).
The relation \(ar=r^{-1}a\) lets us move every occurrence of \(a\) to the left in any word. Since \(a^2=1\), at most one \(a\) remains, and since \(r^n=1\), the exponent of \(r\) may be reduced modulo \(n\).
3. In fact\[ |G_n|=2n. \]
Let \(D_{2n}\) be the symmetry group of a regular \(n\)-gon (for \(n=1,2\), use the same abstract presentation below). Let \(R\) be a rotation of order \(n\) and \(S\) a reflection, so \[ S^2=1,\qquad SRS=R^{-1}. \] Set \(A=S\) and \(B=SR\). Then \[ A^2=B^2=1,\qquad AB=R,\qquad (AB)^n=1. \] Therefore the defining presentation of \(G_n\) gives a surjection \[ G_n\twoheadrightarrow D_{2n}. \] The target has \(2n\) elements, while 2 gives \(|G_n|\le2n\). Hence \(|G_n|=2n\) and \(G_n\cong D_{2n}\).
4. For \(n=4\), the group \(G_4\cong D_8\) has four one-dimensional complex irreducible representations.
Write \[ G_4=\langle r,a\mid r^4=a^2=1,\ ara=r^{-1}\rangle. \] In a one-dimensional representation, the images of \(r\) and \(a\) commute. The relation \(ara=r^{-1}\) therefore forces \[ \rho(r)=\rho(r)^{-1}, \] so \(ho(r)\in\{1,-1\}\). Also \(ho(a)\in\{1,-1\}\). Thus each pair \[ (\rho(r),\rho(a))\in\{\pm1\}^2 \] defines a one-dimensional representation, giving four pairwise nonisomorphic characters. Equivalently, \[ G_4^{\mathrm{ab}}\cong C_2\times C_2. \]
5. There is a two-dimensional irreducible representation\[ \rho(r)= \begin{pmatrix} 0&-1\\ 1&0 \end{pmatrix}, \qquad \rho(a)= \begin{pmatrix} 1&0\\ 0&-1 \end{pmatrix}. \]
These matrices satisfy \[ \rho(r)^4=I,\qquad \rho(a)^2=I,\qquad \rho(a)\rho(r)\rho(a)=\rho(r)^{-1}, \] so they define a representation of \(G_4\). If it had a one-dimensional invariant subspace, that line would be invariant under \(ho(r)\). But \(ho(r)\) has eigenlines over \(\mathbb C\) spanned by \((1,\pm i)\), and \(ho(a)\) interchanges these two lines rather than preserving either one. Hence no nonzero proper subspace is invariant, so the representation is irreducible.
6. These five representations are all the irreducible complex representations of \(G_4\).
For a finite group over \(\mathbb C\), the dimensions \(d_i\) of the irreducible representations satisfy \[ \sum_i d_i^2=|G_4|=8. \] The four one-dimensional representations from 4 contribute \(4\), and the two-dimensional representation from 5 contributes another \(4\). Thus \[ 4\cdot1^2+1\cdot2^2=8, \] so no further irreducible representations exist.