I don't get last two line, I don't know how to prove any of them. if you can give me some hint it would be good
$\Omega_1=$ subgroup generated by all elements of order p
$\Omega^\star_1=\langle x \in G|x^p\in Z(G)\rangle$
I don't get last two line, I don't know how to prove any of them. if you can give me some hint it would be good
$\Omega_1=$ subgroup generated by all elements of order p
$\Omega^\star_1=\langle x \in G|x^p\in Z(G)\rangle$
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I will explain the first two equalities, and leave the rest to you. Let $\Phi = \Phi(G)$.
From the definitions, we have $A \le A^* \le Z(\Phi)$, and they are all abelian. So $d(A_i) = d(A \cap Z_i) \le d(A^* \cap Z_i) \le d( Z(\Phi) \cap Z_i)$, and we want to prove that these are all equalities.
For any abelian $p$-group $Q$, we have $d(Q) = d(\Omega_1(Q))$. So if $d(Z(\Phi) \cap Z_i) = k$, then $Z(\Phi) \cap Z_i$ has an elementary abelian subgroup of rank $k$. But this subgroup lies in $A_i$, so $d(A_i) \ge k$. Hence $d(A_i) = d(A^* \cap Z_i) = d( Z(\Phi) \cap Z_i)$ as claimed.