if G be a finite p-group and G' be isomorphic to zp, what we can say about z(G) by concept of symplectic spaces? is [G:z(G)] a perfect square?
((i take the elementary abelian group G/Z(G) as a vector space on G'but i cant get a proper result.)) please give me some help. tanks
If $G'$ has order $p$ and $G$ is a $p$-group, then we can define a function $b:G\times G \to G'$ by $b(g,h) = [g,h]$. Notice that if $z \in Z(G)$, then $b(zg,h)=b(g,zh)=b(g,h)$ so that $b$ is still well-defined on $G/Z(G) \times G/Z(G)$.
Note that $G' \leq Z(G)$ since $G'$ is a normal subgroup of order $p$ in a $p$-group. Hence $G/Z(G)$ is an abelian $p$-group.
Also note that $b(g,g)$ is the zero (identity) element of $G'$ and $b(g,h)=0$ for all $h \in G$ iff $g \in Z(G)$.
Also $b(g^p,h)=b(g,h)^p$ is the zero element as well, since $G'$ is both central and exponent $p$. Hence $G^p \leq Z(G)$.
This gives the elementary abelian $p$-group $G/Z(G)$ the structure of a non-degenerate symplectic vector space over field of $p$ elements. Since every non-degenrate symplectic space has a basis of hyperbolic pairs, we get that $G/Z(G)$ has even dimension, so its order is a square.