For any group \(G\) we define \(\Omega(G)\) to be the image of the group homomorphism \(\rho:G\rightarrow\Aut(G)\) where \(\rho\) maps \(g\in G\) to the conjugation automorphism \(x\mapsto gx\inverseof{g}\). Starting with a group \(G_0\), we define \(G_1=\Omega(G_0)\) and \(G_{i+1}=\Omega(G_i)\) for all \(i\geq 0\). If \(G_0\) is of order \(p^e\) for a prime \(p\) and integer \(e\geq 2\), prove that \(G_{e-1}\) is the trivial group.
Solution
1. For every group \(G\),\[ \Omega(G)\cong G/Z(G). \]
The homomorphism \[ \rho:G\longrightarrow \operatorname{Aut}(G), \qquad \rho(g)(x)=gxg^{-1}, \] has image \(\Omega(G)\) by definition. Its kernel consists exactly of the elements commuting with every element of \(G\), hence is \(Z(G)\). The first isomorphism theorem gives \[ \Omega(G)\cong G/Z(G). \]
2. If \(H\) is a finite \(p\)-group of order \(p^m\), then either \(\Omega(H)=1\) or\[ |\Omega(H)|\le p^{m-2}. \]
By 1, \(\Omega(H)\cong H/Z(H)\). Since \(H\) is a finite \(p\)-group, \(Z(H)\) is nontrivial whenever \(H\) is nontrivial, so \(|H/Z(H)|\le p^{m-1}\).
It remains to exclude the case \(|H/Z(H)|=p\). If \(H/Z(H)\) were cyclic, say generated by \(hZ(H)\), then every element of \(H\) would have the form \(h^r z\) with \(z\in Z(H)\). Thus any two elements \(h^r z\) and \(h^s z'\) commute, so \(H\) would be abelian. Then \(Z(H)=H\), making \(H/Z(H)\) trivial, a contradiction. Therefore a nontrivial quotient \(H/Z(H)\) cannot have order \(p\). Since its order is a power of \(p\), a nontrivial \(\Omega(H)\) has order at most \(p^{m-2}\).
3. For every \(i\ge0\), either \(G_i=1\) or\[ |G_i|\le p^{e-2i}. \]
We use induction on \(i\). For \(i=0\), \(|G_0|=p^e\). Suppose the claim holds for \(i\). If \(G_i=1\), then \(G_{i+1}=\Omega(G_i)=1\). Otherwise \(G_i\) is a finite \(p\)-group of order at most \(p^{e-2i}\). Applying 2, \[ G_{i+1}=1 \quad\text{or}\quad |G_{i+1}|\le p^{e-2i-2}=p^{e-2(i+1)}. \] This proves the induction step.
4. The group \(G_{e-1}\) is trivial.
If \(G_{e-1}\) were nontrivial, then 3 would give \[ 1<|G_{e-1}|\le p^{e-2(e-1)}=p^{2-e}\le1, \] because \(e\ge2\). This is impossible. Hence \(G_{e-1}=1\).