Center of a split metabelian group of order $p^nq$ is direct summand.

Question

Let $p$ and $q$ be odd primes such that such that $q \ | \ p-1.$ Suppose we have a group $G$ which is a split extension of the cyclic group $B$ of order $q$ by a finite abelian $p$-group $A.$ Can we say that $Z(G)$ is a direct summand of $G?$

I have tried up to some extent but I am stuck.

Suppose $B = \langle b \rangle.$ Then as $Z(G) \subseteq A,$ we have that all the elements of $A$ which commute with $b$ are the only central elements of $G.$ If $Z(G)$ is non-trivial, then we only need to prove that it is a direct summand of $A.$ For that assume that $A = \langle a_1 \rangle \times \langle a_2 \rangle \times \ldots \times \langle a_n \rangle$ with $o(a_i) = p^i.$ If for some $i,$ $a_i^{p^k}$ commutes with $b,$ then I am trying to prove that $a_i$ also commutes with $b.$

Any comments would be highly appreciated!

What do you mean by and extension of $B$ by $A$? There are two possible interpretations, so you should say which one you mean. — Derek Holt, Jan 30 '20 at 12:21

score 3 · Accepted Answer · answered Jan 30 '20 at 13:55

3

This follows from the more general result that if a finite $p'$-group $B$ acts on a finite abelian $p$-group $A$ then $A = C_B(A) \times [B,A]$.

There is a proof in (24.6) of Aschbacher's book for example.

answered Jan 30 '20 at 13:55

Derek Holt

96,726

What if $G$ is nilpotent but not abelian? – the_fox Jan 30 '20 at 14:07
Ah, sorry. I overlooked that $A$ is abelian. – the_fox Jan 30 '20 at 15:24
If $A$ is a not necessarily abelian $p$-group then we have $A=C_B(A)[B,A]$ but they can intersect. – Derek Holt Jan 30 '20 at 15:27
Thank you very much @ Derek Holt – math seeker Jan 31 '20 at 07:51

Center of a split metabelian group of order $p^nq$ is direct summand.

1 Answers1