# Lift a group action from a quotient

Let $p$ be a rational prime and $H$ be a finite cyclic group of prime order $l$ prime to $p$, i.e. $(l,p) = 1$. Let $G$ be a finite abelian group of $p$-power order.

If I can write an (abelian) group $\Delta$ as a direct product like $$\Delta = G \times H$$ then $H$ acts trivially on $G$. Now I know that $G$ is a quotient group, say $G = G_1 / N$, via a normal subgroup $N$ of a finite abelian group $G_1$ of exponent $p$.

What can I say about the action of $H$ on $G_1$?

I need to find the representations of the group $H$, that acts on (the $\mathbb{F}_p$ vector space) $G_1$, but I don't know about the action.

Background The ultimative goal is described in this thread. All of the above groups are in fact Galois groups of certain field extensions and I'm interested in the link between Galois theory and representation theory.

Module formulation Since $H$ acts trivially on the finite abelian $p$-group $G$, $G$ becomes a trivial $\mathbb{F}_{p^n}[H]$ module. My question is therefore analogous to: Does $H$ act trivially on a $\mathbb{F}_p[H]$-submodule of $G_1$? In other words: Does the $\mathbb{F}_p[H]$-module $G_1$ contain a trivial $\mathbb{F}_p[H]$-submodule? (I'm not sure about this since $G$ is just a quotient of $G_1$ by the normal subgroup $N$) Notice that $\mathbb{F}_p[H]$ is semisimple by Maschke's theorem.

Thank you :-)

• Added a module formulation. – BIS HD Oct 24 '13 at 14:43
• Why do you start out only saying $G$ has $p$-power order; why not state up front it has exponent $p$? What does $G\times H$ have to do with anything; why not just say $H$ acts on $G$ trivially? How are you justified in talking about "the" action of $H$ on $G_1$; in general won't there by numerous actions of $H$ on $G_1$ that stabilize $N$ and act trivially on $G_1/N$? – whacka Feb 25 '15 at 23:38