# Showing $d(\frac{Z_2(G)}{Z(G)})=d(G)d(Z(G))$ if and only if $d(G)=2$.

I am tring to proves following lemma

Let $G$ be a finite p-group of coclass 3 and nilpotency class greater than 3 then $d(\frac{Z_2(G)}{Z(G)})=d(G)d(Z(G))$ if and only if $d(G)=2$

correct me if i'm wrong
since coclass=n-c, n is from $p^n$ the order of G and c is nilpotency class then in central serie for some i (max 3) we have $\frac{Z_{i+1}(G)}{Z_i(G)}$ order is $p^2$ or $p^3$ or $p^4$
since $d(G)=2$ then $d(Z(G))\le2$
can any one help me please what I'm missing
and if above lemma hold how can i show that $\frac{Z_2(G)}{Z(G)}$ is isomorphic to an elementary abelian p-group of order $p^2$ ?

• One does not "solve" a lemma; one "proves" it. – Shaun May 27 '18 at 19:01
• I don't believe it. It is possible to have $d(G)=2$, $d(Z_2(G)/Z(G))=1$ and $d(Z(G))=2$, and in $\mathtt{SmallGroup}(128,48)$. – Derek Holt May 27 '18 at 19:13