Let \(G\) be the group of order \(8\) defined by the relations \[ a^4=1\quad\text{and}\quad a^2=b^2\quad\text{and}\quad b^{-1}ab=a^3. \]
Show that \(ab=b^3a\) and that every element of \(G\) is of the form \(b^k\) or \(b^ka\) where \(k=0,\ldots,3\).
Given that the conjugacy classes of \(G\) are \[\begin{align*} C_1&=\{1\},\\ C_2&=\{b^2\},\\ C_3&=\{b,b^3\},\\ C_4&=\{a,b^2a\},\\ C_5&=\{ba,b^3a\}, \end{align*}\]
Show that \(H=\{1,b^2\}\) is a normal subgroup of \(G\) for which \(G/H\) is isomorphic to \(Z_2\times Z_2\).
Give the character character table for the lifting of the \(4\) linear characters of \(G/H\) to \(G\).
Find the complete character table for \(G\).
Solution
1. The relations imply\[ ab=b^3a, \]and every element of \(G\) has the form \(b^k\) or \(b^ka\) with \(0\le k<4\).
From \[ b^{-1}ab=a^3 \] we get \[ ab=ba^3. \] Also \(a^2=b^2\). Since \[ b^{-1}a^2b=(b^{-1}ab)^2=a^6=a^2, \] the element \(a^2=b^2\) is central. Hence \[ ba^3=ba^2a=bb^2a=b^3a, \] so \(ab=b^3a\).
Using this relation, every word in \(a,b\) can be moved into the form \(b^ka^\varepsilon\) with \(\varepsilon\in\{0,1\}\): replace every occurrence of \(a^2\) by \(b^2\) and move the remaining \(a\) to the right. Since \(b^4=(b^2)^2=(a^2)^2=a^4=1\), we may take \(0\le k<4\). The problem states that \(|G|=8\), so these eight normal forms are precisely all elements of \(G\).
2. The subgroup\[ H=\{1,b^2\} \]is normal, and\[ G/H\cong C_2\times C_2. \]
The element \(b^2=a^2\) is central by 1, so \(H\trianglelefteq G\). In the quotient, \[ (aH)^2=a^2H=H, \qquad (bH)^2=b^2H=H. \] Moreover \(G/H\) has order \(4\). It is not cyclic: every nonidentity element has order \(2\). Hence \[ G/H\cong C_2\times C_2. \]
3. The four linear characters lifted from \(G/H\) have the following values on the five conjugacy classes:\[ \begin{array}{c|rrrrr} & C_1 & C_2 & C_3 & C_4 & C_5\\ \hline \chi_0&1&1&1&1&1\\ \chi_1&1&1&-1&1&-1\\ \chi_2&1&1&1&-1&-1\\ \chi_3&1&1&-1&-1&1 \end{array}. \]
Every linear character of \(C_2\times C_2\) is determined by independently sending the two generators \(bH\) and \(aH\) to \(\pm1\). All such characters are trivial on \(H\), so they take value \(1\) on \(C_1\) and \(C_2\). The classes \(C_3,C_4,C_5\) map respectively to \(bH,aH,baH\), giving exactly the four rows displayed above.
4. There is a two-dimensional representation \(\rho\) of \(G\) given by\[ \rho(a)= \begin{pmatrix} i&0\\ 0&-i \end{pmatrix}, \qquad \rho(b)= \begin{pmatrix} 0&1\\ -1&0 \end{pmatrix}. \]Its character has values\[ \chi_4=(2,-2,0,0,0) \]on \(C_1,\ldots,C_5\).
Both matrices square to \(-I\), so \[ \rho(a)^2=\rho(b)^2=-I, \qquad \rho(a)^4=I. \] A direct multiplication gives \[ \rho(b)^{-1}\rho(a)\rho(b)=\rho(a)^{-1}=\rho(a)^3, \] so the defining relations of \(G\) are satisfied.
The traces are \[ \operatorname{tr}I=2, \qquad \operatorname{tr}(-I)=-2, \] and each of \(\rho(a),\rho(b),\rho(ba)\) has trace \(0\). Since character values are constant on conjugacy classes, this gives the stated row.
5. The representation \(\rho\) is irreducible, and the complete character table of \(G\) is\[ \boxed{ \begin{array}{c|rrrrr} & C_1 & C_2 & C_3 & C_4 & C_5\\ \text{class size}&1&1&2&2&2\\ \hline \chi_0&1&1&1&1&1\\ \chi_1&1&1&-1&1&-1\\ \chi_2&1&1&1&-1&-1\\ \chi_3&1&1&-1&-1&1\\ \chi_4&2&-2&0&0&0 \end{array}.} \]
For \(\chi_4\), \[ \langle\chi_4,\chi_4\rangle =\frac1{8}(4+4)=1, \] so \(\rho\) is irreducible. The four linear characters are irreducible automatically. Their degree squares sum to \[ 1+1+1+1+4=8=|G|, \] so these five irreducibles exhaust all irreducible representations of \(G\).