# Sufficient condition for complementation of abelian normal subgroup

Suppose that we are given a finite $$p$$-group $$G$$ and an abelian normal subgroup $$A$$ of $$G$$. The question I have is whether any sufficient conditions are known for $$A$$ to have a complement in $$G$$. From briefly looking around a bit, I have found nothing on this topic.

I would be most interested if there were a necessary and sufficient condition for $$A$$ to have a complement in $$G$$, but I am fairly confident that no such condition is known.

The only thing in this line that I am aware of is a theorem of Gaschütz which works more generally:

Ιf $$A$$ is an abelian normal subgroup of the finite group $$G$$ and $$A \cap \Phi(G)=1$$, then $$A$$ has a complement in $$G$$.

Of course, if $$G$$ is assumed to be a $$p$$-group then only an elementary abelian subgroup $$A$$ can satisfy Gaschütz's condition. It follows easily from Gaschütz's theorem that if $$A \leq \Omega_1(Z(G))$$, that is, if $$A$$ is central and elementary abelian, then $$A$$ has a complement in $$G$$ if and only if $$A \cap \Phi(G)=1$$.

It is perhaps worth pointing out that if $$G$$ is abelian, then a necessary and sufficient condition for $$A$$ to have a complement in $$G$$ is for $$A$$ to be a pure subgroup of $$G$$. This, in turn, admits the equivalent description: $$A$$ is pure in $$G$$ if and only if the Frattini series of $$A$$ is the intersection of $$A$$ with the Frattini series of $$G$$.

I would also be interested in partial results of this kind, i.e. by assuming something extra about either $$G$$ or $$A$$. Particular cases of interest would be $$A$$ central or $$A$$ cyclic.

Perhaps it would be useful if I added a little thing I have found. Of course, this lemma can only work if $$A$$ is, for instance, non-cyclic. In fact, it seems to me that the case $$A$$ cyclic is already difficult because there are simply not enough subgroups to work with.

Lemma: Let $$G$$ be a finite nilpotent group and let $$A$$ be normal in $$G$$. Suppose that $$A=CD$$, where $$C$$, $$D$$ are both normal in $$G$$ and that $$(C \cap D) \cap \Phi(G) = 1$$. Then $$A$$ is complemented in $$G$$ if and only if $$A/C$$ is complemented in $$G/C$$ and $$A/D$$ is complemented in $$G/D$$.

Proof. Suppose first that $$A$$ is complemented in $$G$$ and let $$N \unlhd G$$ with $$N \leq A$$. We argue that if $$H$$ complements $$A$$ in $$G$$, then $$HN/N$$ complements $$A/N$$ in $$G/N$$. Since $$N \leq A$$, we have by Dedekind's lemma that $$A \cap HN = (A \cap H)N = N,$$ thus $$(A/N) \cap (HN/N) = 1$$. On the other hand, $$A(HN) = AH = G$$ thus $$(A/N)(HN/N) = G/N,$$ as wanted. Applying this with $$N=C$$ and $$N=D$$ proves one direction.

Now we assume that both $$A/C$$ and $$A/D$$ are complemented in $$G/C$$ and $$G/D$$ respectively and we work to show that $$A$$ has a complement in $$G$$. Suppose then that $$H/C$$ is a complement of $$A/C$$ in $$G/C$$ and $$K/D$$ is a complement of $$A/D$$ in $$G/D$$. Let us denote $$X := H \cap K$$ and $$I :=C \cap D$$.

We begin by observing that $$I \leq X$$. Since $$I \cap \Phi(G) = 1$$ by hypothesis, we have $$\Phi(I) \leq I \cap \Phi(G) = 1,$$ thus $$\Phi(I) = 1$$ and it follows that $$I$$ is abelian. Moreover, $$I$$ is a normal subgroup of $$G$$.

Now $$I \unlhd X$$, $$I$$ is abelian, and $$I \cap \Phi(X) \leq I \cap \Phi(G) = 1,$$ where $$\Phi(X) \leq \Phi(G)$$ holds because $$G$$ is nilpotent. It follows by Gaschütz's theorem mentioned above that $$I$$ has a complement in $$X$$ and we call that complement $$R$$. We will argue that $$R$$ complements $$A$$ in $$G$$ and our first observation to show this is that $$A \cap R = A \cap X \cap R = ((A \cap H) \cap (A \cap K)) \cap R = C \cap D \cap R = I \cap R = 1.$$ It thus only remains to show that $$AR=G$$. Since $$AX = A(IR) = AR$$, it will suffice to show that $$G = AX$$.

We have $$G/C = (H/C)(A/C)$$, so $$G = HA = HCD = HD,$$ where the final equality holds because $$C \leq H$$. Thus $$G = HD$$, and similar reasoning yields $$G = KC$$. Now $$G = HD$$ and $$D \leq K$$, so $$K = (H \cap K)D = XD$$ by Dedekind's lemma. Then $$G = KC = XDC = XA.$$ $$\blacksquare$$

Note: The same conclusion can be reached without assuming that $$G$$ is nilpotent and assuming instead that either $$A$$ or $$I$$ is abelian. The crucial point is to ensure that $$I$$ has a complement in $$X$$.

• A different kind of partial result yields a supplement $B$, i.e., $AB=G$ and $A\cap B<A$. For example, if $[A,G]=[G,G]\cap A$ and $A/[A,G]$ is complemented in the abelian group $G/[G,G]$, then $B$ exists satisfying $AB=G$ and $A\cap B=[A,G]$. Dec 6, 2019 at 19:28