Quotient of a minimal non abelian $p$-group by its center

Question

I am trying to solve the following question:

Let $P$ be a nonabelian finite $p$-group in which every proper subgroup is abelian. Show that $P/Z(P) \simeq C_p \times C_p .$

Since $P$ is non-abelian I know two things

1. $p^2$ divides $[P:Z(P)]$
2. $P/Z(P)$ is not cyclic.

How can I show that $[P:Z(P)]=p^2$?

Answer 1

Perhaps a more direct argument: Let $M$ and $N$ be distinct maximal subgroups of the minimal non-abelian $p$-group $G$. Then $M$ and $N$ are abelian and normal in $G$. Hence, $G=MN$ and $M\cap N\le Z(G)$. It follows that $$|G:Z(G)|\le|G:M\cap N|=|G:M||M:M\cap N|=p|MN:N|=p^2.$$ Since $G/Z(G)$ is non-cyclic, we have $|G:Z(G)|=p^2$.

Answer 2

Lemma 1. Let $G$ be a group, and let $M$ and $N$ be subgroups of $G$. Then $Z(M)\cap Z(N)\leq Z(\langle M,N\rangle)$.

Proof. If $x\in Z(M)\cap Z(N)$, then $M,N\leq C_G(\{x\})$. Therefore, $\langle M,N\rangle\leq C_G(\{x\})$, so $x\in Z(\langle M,N\rangle)$, given that it lies in $M\cap N$. $\Box$

Lemma 2. Let $G$ be a group, and let $M$ and $N$ be distinct maximal subgroups of $G$. Then $Z(M)\cap Z(N)\leq Z(G)$.

Proof. Follows from Lemma 1, since $\langle M,N\rangle = G$. $\Box$

Lemma 3. Let $G$ be a finite nonabelian group, and let $M$ be a maximal subgroup of $G$. If $M$ is abelian, then $Z(G)\leq M$.

Proof. Since $M\leq MZ(G)\leq G$, we have either $M=MZ(G)$ or $MZ(G)=G$. If $MZ(G)=G$, then because $M$ is abelian, and $Z(G)$ centralizes $M$, we have $G$ abelian. Therefore, $MZ(G)=M$ so $Z(G)\leq M$. $\Box$

Lemma 4. Let $G$ be a finite nonabelian $p$-group such that every proper subgroup is abelian. If $M$ and $N$ are distinct maximal subgroups of $G$, then $Z(G)=M\cap N$.

Proof. From Lemma 2, we know that $M\cap N = Z(M)\cap Z(N) \leq Z(G)$. From Lemma 3, we know that $Z(G)\leq M\cap N$. This proves equality. $\Box$

Corollary. Let $G$ be a finite nonabelian $p$-group such that every proper subgroup is abelian. Then $\Phi(G)\leq Z(G)$, where $\Phi(G)$ is the Frattini subgroup of $G$.

Proof. Since $G$ is nonabelian, it has at least two distinct maximal subgroups $M_1\neq M_2$. So $$\Phi(G) = \cap_{M\text{ maximal}}M \leq M_1\cap M_2 = Z(G).\quad \Box.$$

Let $P$ be a nonabelian $p$-group with the property that every proper subgroup of $P$ is abelian. Since $P/\Phi(P)$ is an elementary abelian $p$-group, and $\Phi(P)\leq Z(P)$, it follows that $P/Z(P)$ is an elementary abelian $p$-group.

Any two distinct maximal subgroups of $P$ have intersection equal to $Z(P)$, so any two maximal subgroups of $P/Z(P)$ have trivial intersection, by the Lattice Isomorphism Theorem.

View $P/Z(P)$ as a vector space over $\mathbb{F}_p$. The previous paragraph tells us that the intersection of two subspaces of codimension $1$ is always trivial. But that intersection has codimension $2$: indeed, if $\mathbf{V}$ is a vector space of dimension $1$, and $\mathbf{W}_1,\mathbf{W}_2$ are distinct of dimension $n-1$, then $$\begin{align*} n&= \dim(\mathbf{V}) = \dim(\mathbf{W}_1+\mathbf{W}_2)\\ &= \dim(\mathbf{W}_1)+\dim(\mathbf{W}_2) - \dim(\mathbf{W}_1\cap\mathbf{W}_2)\\ &= (n-1)+(n-1) - \dim(\mathbf{W}_1\cap\mathbf{W}_2)\\ &= 2n-2 - \dim(\mathbf{W}_1\cap\mathbf{W}_2). \end{align*}$$ Therefore, $\dim(\mathbf{W}_1\cap\mathbf{W}_2) = n-2 = \dim(\mathbf{V}))-2$.

Since the intersection of two subspaces of codimension $1$ has codimension $2$, and in this case it has dimension $0$, it follows that $\dim(\mathbf{V})=2$. That is, $P/Z(P)$ has order $p^2$. As it cannot be cyclic, we must have $Z/P(Z)\cong C_p\times C_p$, as claimed.