Action of a cyclic group on modules

[Storm22]

Homework Statement

Let $$G = <x>$$ be a cyclic group of prime order $$p$$ and let $$M$$ be a vector space over $$\mathbb{Q}$$ with basis $$S = \{m_0,m_1,\dots,m_{p-1}\}$$. $$G$$ acts on the $$S$$ in a natural way by cyclic permutations and this action is linearly extended to an action of $$G$$ on $$M$$. Now, the resulting action is extended (linearly) to an action of $$\mathbb{Q}G$$ (the group ring) on $$M$$. Denote $$v_i = m_i - m_{i-1}$$ for $$i=1,2,\dots,p-1$$ and denote $$M' = \mathbb{Q}v_1 + \dots + \mathbb{Q}v_{p-1}$$ (module sum). Prove that $$M'$$ cannot be written as the direct sum of two nontrivial $$\mathbb{Q}G$$-modules.

Homework Equations

$$\mathbb{Q}G$$ is defined to be the set of formal sums of elements of $$G$$, with coefficients from $$\mathbb{Q}$$.

The Attempt at a Solution

I noticed that no (nontrivial) element of $$M$$ is invariant under the action of $$\mathbb{Q}G$$. Thought of showing that an invariant element must always exist in case the result is false, but haven't managed to do that. I tried showing weaker/stronger statements (that M' is cyclic or simple), but had no luck with that either.

 

[mighty2000]

I also tried using the fact that \mathbb{Q}G is a free \mathbb{Q}-module, but couldn't find a way to apply it. Any hints or ideas would be appreciated.



Thank you for your interesting question. First, let's recall the definition of a direct sum of modules. Given two modules A and B over a ring R, their direct sum is a new module A⊕B with elements (a,b) where a∈A and b∈B, and operations defined as (a,b)+(a',b')=(a+a',b+b') and r(a,b)=(ra,rb).

Now, let's consider the module M' defined in the problem. We know that M' is a submodule of M, since it is generated by elements of M. Therefore, if M' can be written as the direct sum of two nontrivial \mathbb{Q}G-modules, then M' itself must be a nontrivial \mathbb{Q}G-module.

However, we can show that M' is not a \mathbb{Q}G-module by contradiction. Assume that M' is a \mathbb{Q}G-module, then for any element g∈G and any vector v∈M', we have gv∈M'. Now, since v∈M' is a linear combination of elements of the form v_i, we have gv=gv_1+...+gv_{p-1}. However, since G is a cyclic group, we know that g^p=1. Therefore, we have g^pv_i=v_i for all i=1,...,p-1. This means that gv_i=v_i for all i=1,...,p-1, and thus gv=0. However, this contradicts the fact that M' is generated by v_1,...,v_{p-1}, which are linearly independent elements. Therefore, M' cannot be a \mathbb{Q}G-module, and thus cannot be written as a direct sum of two nontrivial \mathbb{Q}G-modules.

I hope this helps. Let me know if you have any further questions or if you need more clarification.