Give an example of a group \(G\) such that the center of \(G\) modulo its center is non-trivial. Give an example of a group \(H\) such that the groups \(H\), \(H'\), \(H''\) and \(H'''\) are all distinct. (The derived group \(H'\) of a group \(H\) is the subgroup generated by commutators, or equivalently the smallest subgroup such that the quotient by this subgroup is an abelian group.)
Solution
For the first example, let \[ G \coloneqq \langle r,s\mid r^4=s^2=1,\ srs=r^{-1}\rangle, \] the dihedral group of order \(8\).
1. The center of \(G\) is\[ Z(G)=\{1,r^2\}. \]
The element \(r^2\) commutes with \(r\), and \[ sr^2s =(srs)^2 =r^{-2} =r^2, \] so \(\{1,r^2\}\subseteq Z(G)\).
The rotations \(r\) and \(r^3\) are not central because \(srs=r^{-1}\ne r\). A reflection \(r^ks\) does not commute with \(r\): \[ r(r^ks)=r^{k+1}s, \qquad (r^ks)r=r^{k-1}s, \] and these are unequal because \(r^2\ne1\). Thus no other element is central.
2. The center of \(G/Z(G)\) is nontrivial; indeed,\[ G/Z(G)\cong C_2\times C_2. \]
The quotient has order \(4\). The cosets of both \(r\) and \(s\) have order \(2\), and modulo \(Z(G)\) the relation \[ srs=r^{-1} \] becomes \[ (sZ(G))(rZ(G))(sZ(G)) =r^{-1}Z(G) =rZ(G), \] because \(r^{-1}=r^3=r\,r^2\) and \(r^2\in Z(G)\). Hence the quotient is abelian and noncyclic, so it is \(C_2\times C_2\). Therefore \[ Z(G/Z(G))=G/Z(G)\ne1. \]
3. For the second example, take\[ H=S_4. \]Then\[ H'=A_4. \]
The sign map \[ S_4\longrightarrow\{\pm1\} \] has abelian target and kernel \(A_4\), so \(S_4'\subseteq A_4\). On the other hand, \[ [(12),(23)] \] is a \(3\)-cycle. Since \(S_4'\) is normal, it contains every conjugate of this \(3\)-cycle, hence every \(3\)-cycle. The group \(A_4\) is generated by \(3\)-cycles, so \(A_4\subseteq S_4'\). Thus \(S_4'=A_4\).
4. One has\[ H''=A_4'=V_4, \]where\[ V_4 = \{1,(12)(34),(13)(24),(14)(23)\}. \]
The subgroup \(V_4\) is normal in \(A_4\), and \[ A_4/V_4 \] has order \(3\), hence is abelian. Therefore \(A_4'\subseteq V_4\).
With the convention \([a,b]=aba^{-1}b^{-1}\), direct computation gives \[ [(123),(124)]=(12)(34). \] Thus \(A_4'\) contains a nonidentity element of \(V_4\). Since \(A_4'\) is normal in \(A_4\), it also contains the \(A_4\)-conjugates of \((12)(34)\), which are the other two nonidentity elements of \(V_4\). Hence \(V_4\subseteq A_4'\), and equality follows.
5. One has\[ H'''=V_4'=1. \]
The Klein four-group \(V_4\) is abelian, so its derived subgroup is trivial.
6. The four groups\[ \boxed{ S_4,\quad A_4,\quad V_4,\quad 1 } \]are all distinct.
Their orders are respectively \[ 24,\quad12,\quad4,\quad1. \] Together with steps <1>3–<1>5, this gives the required derived series.
7. Q.E.D.
Step 2 supplies the required group \(G\), and step 6 supplies the required group \(H\).