groups of order $p^5$ and exponent p We know that a group of order $p^5$ is an extra special group. How we can show it doesn't have any abelian subgroup of order $p^4$? Also what is the presentation of this group if its exponent is equal to $p$ ?
If $P$ is an extraspecial non-Abelian $p$-group of order $p^{5},$ then $Z(P) = [P,P] = \Phi(P)$ has order $p,$ so is cyclic, say equal to $\langle z \rangle.$ Then $P/Z(P)$ is an elementary Abelian group of order $p^{4},$ and may be viewed as a $4$-dimensional vector space $V$ over $F,$  the field with $p$-elements.

The vector space $V$ admits a well defined non-degenerate alternating $F$-bilinear form $\langle, \rangle :V \times V \to F$ by setting $\langle xZ(P),yZ(P) \rangle = j1_{F}$ if $[x,y] = z^{j},$ (where $0 \leq j \leq p-1$).

If $A$ is an Abelian subgroup of $P$ containing $Z(P),$ then $A/Z(P) \leq \left( A/Z(P) \right)^{\perp},$ so $A/Z(P)$ is at most $2$-dimensional, as the form is non-degenerate.
Hence $|A| \leq p^{3}.$ 

An alternative proof may be constructed with some knowledge of character theory. If $P$ had an Abelian subgroup $A$ of order $p^{4},$ then by Clifford's theorem, every irreducible character of $P$ of degree greater than $1$ is induced from a degree $1$ irreducible character of $A,$ so has degree at most $p.$ However, $P$ has $p-1$ complex irreducible characters of degree $p^{2}$ and $p^{4}$ irreducible characters of degree $1,$ so there can be no such $A.$
Whether, in the above example, the equation C_P(x)=Z(P) , (x is not in Z(P)), holds or not?
In fact, I am looking for a 2-group such that P/Z(P) is an elementary abelian, but
C_P(x)≠Z(P) .