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Show that a finite non-abelian $p$-group cannot split over its center.

I'd be happy for some clues.

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Write $P = Q \times Z(P)$. How are $Z(Q)$ and $Z(P)$ related? How big is $Z(Q)$? – Jack Schmidt Jul 14 '12 at 19:13
@JackSchmidt: I know that splitting gives a direct sum in abelian categories, but I've never seen a corresponding result in the category of groups. Are you asserting that if we have maps $H\to G \to H$ with composite the identity map, then $G=H\times K$ for some group $K$? If so, is there an easy proof? Or is that just what it means to spilt in group theory? – Aaron Jul 14 '12 at 23:22
@Aaron: I think Jack just misspoke. "Splitting over $K$" means that $G$ is a semidirect product with $K$ normal. – Arturo Magidin Jul 14 '12 at 23:48
@ArturoMagidin: Ahh, thanks. – Aaron Jul 15 '12 at 1:36
@Aaron: Splitting over a central subgroup is always a direct product. In general it is only semi. – Jack Schmidt Jul 15 '12 at 2:12
up vote 2 down vote accepted

This definition is not very common, so it may be worth mentioning here:

Definition. Let $G$ be a group. A subgroup $K$ of $G$ is said to be co-central in $G$ if $G=Z(G)K$.

Theorem. Let $G$ be a group. If $K$ is co-central in $G$, then $Z(K)\subseteq Z(G)$.

Proof. Let $z\in Z(K)$, and $x\in G$. We need to show that $zx=xz$. Since $G=Z(G)K$, there exists $a\in Z(G)$ and $k\in K$ such that $x=ak$. Then $$\begin{align*} zx &= zak\\ &= azk &&\text{(since }a\in Z(G)\text{)}\\ &= akz &&\text{(since }z\in Z(K)\text{ and }k\in K\text{)}\\ &= xz. \end{align*}$$ Thus, $z\in Z(G)$, as claimed. $\Box$

Now assume that we can write a $p$-group $P$ as $P=Z(P)H$ with $H$ a subgroup. Then by the lemma, $Z(H)\subseteq Z(P)$. If $P$ were split over $Z(P)$, then we would have $Z(P)\cap H = \{e\}$, hence $Z(H)=\{e\}$.

Why is that a very big problem for $H$, which can only be solved if $H=\{e\}$?

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+1 Nice explanation and even nicer theorem! – DonAntonio Jul 15 '12 at 2:05

Suppose $P=Z(P)H$ with $1<H<P$ nontrivial and $H\cap Z(P)=1$. Two questions:

  • Can $Z(H)\subseteq H$ and $Z(P)$ intersect nontrivially?
  • Can $Z(H)=1$? Observe that $H$ is a $p$-group.

Show that every element of $Z(P)Z(H)$ commutes with every element of $Z(P)H=P$, and therefore derive the containment $Z(P)Z(H)\subseteq Z(P)\implies Z(H)\subseteq Z(P)$, contra trivial intersection.

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