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Let $G$ be an extraspecial group of order $p^{2n+1}$. Then
$$
cp(G)=\frac{p^{2n}+p-1}{p^{2n+1}}=\frac1p+\frac{p-1}{p^{2n+1}}.
$$
All maximal abelian subgroupsubgroups have index $p^n$ in $G$. Hence, the answer is `not'.
Let $G$ be an extraspecial group of order $p^{2n+1}$. Then
$$
cp(G)=\frac{p^{2n}+p-1}{p^{2n+1}}=\frac1p+\frac{p-1}{p^{2n+1}}.
$$
All maximal abelian subgroup have index $p^n$ in $G$. Hence, the answer is `not'.
Let $G$ be an extraspecial group of order $p^{2n+1}$. Then
$$
cp(G)=\frac{p^{2n}+p-1}{p^{2n+1}}=\frac1p+\frac{p-1}{p^{2n+1}}.
$$
All maximal abelian subgroups have index $p^n$ in $G$. Hence, the answer is `not'.
Let $G$ be an extraspecial group of order $p^{2n+1}$. Then
$$
cp(G)=\frac{p^{2n}+p-1}{p^{2n+1}}=\frac1p+\frac{p-1}{p^{2n+1}}.
$$
All maximal abelian subgroup have index $p^n$ in $G$. Hence, the answer is `not'.