$|C(E):C(E)\cap C(Z(U))|=1$ or $p$

Question

Lemma 2.4 from Robert Griess' paper "Elementary abelian $p$-subgroups of algebraic groups" states:

(i) Let $T$ be a finite $p$-group whose Frattini subgroup is cyclic and central. Then $T'$ has order $1$ or $p$ and there are subgroups $X, Y$ such that $T$ = $X\circ Y$ where $X$ is extraspecial and $Y$ has an abelian maximal subgroup and $\Omega_1(Y)$ is elementary abelian.

(ii) If $T/T'$ is elementary abelian, $Y$ is of the form $p^r$ or $\mathbb{Z}_{p^2} \times p^r$.

The proof for (i):

Notice that $T'$ has order 1 or $p$. The abelian case is trivial, so we assume that $T'$ has order $p$. Let $U \ge T'$ satisfy $U/T'=\Omega_1(T/T')$. Let $E$ be an extraspecial subgroup of $U$ such that $U=Z(U)E$ where $Z(U)$ denotes the center. Then, $[E,T]=T'=Z(E)$ implies that $T=C(E)E$ where $C(E)$ denotes the centralizer. Since $T/U$ is cyclic and $T'$ has order $p$, we see that $|C(E):C(E)\cap C(Z(U))|=1$ or $p$. Since the Frattini subgroup of $T$ is cyclic, the same is true for subgroups and quotients, whence $C(E)\cap C(Z(U))$ is central-by-cyclic, hence abelian. Take $X=E$ and $Y=C(E)$.

Question: I don't really understand the following: Since $T/U$ is cyclic and $T'$ has order $p$, we see that $|C(E):C(E)\cap C(Z(U))|=1$ or $p$. Did he use some kind of order formula for centralizers or something?

Any help would be appreciated.

Answer 1

One possible argument is to compare this with $T/C(Z(U))$ as follows. We have $$C(E)/C(E)\cap C(Z(U))\cong C(E)C(Z(U))/C(Z(U)) \leqslant T/C(Z(U))$$ Now $T/U$ is cyclic, so $T/C(Z(U))$ is cyclic. It acts faithfully on $Z(U)$ by conjugation, and if $x$ is an element of $T$ whose image is a generator of $T/C(Z(U))$ then $[x,Z(U)]\leqslant T'$ (since $T/T'$ is abelian), so $x^p$ centralises $Z(U)$ since $T'$ has order $p$. Thus $x^p\in C(Z(U))$, which implies that $T/C(Z(U))$ has order $1$ or $p$.