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If $G$ is an abelian group with cyclic subgroup $H$ and $(\rho,V)$ is a (permutation) representation of $G$. Then I can form a representation of $H$ by considering the composition $\lambda=\rho\circ\iota:H\to GL(V)$ where $\iota:H\to G$ is the inclusion map. Now if I let $V^{H}\subset V$ be the subspace of $V$ fixed by $H$ (the image of the projection map $\frac{1}{|H|}\sum_{h\in H}\lambda(h)$) then what can I say about the dimension of $V^{H}$? Can it be understood in terms of the dimension of $V$, the order of $H$, and the order of $G$?

This seems intimately related to the following question. Are there versions of the orbit-stabilizer theorem and/or Burnside lemma in which dimensions (rather than cardinalities) appear and that can be applied in the theory of linear representations?

I'm particularly interested in the case that $(\rho,V)$ is the regular representation modulo the trivial representation.

As to the broader context of this question. Aside from being a purely academic question that I'd like to know the answer to, I was motivated to ask it when attempting to understand an answer provided to the following question on MathOverlow:

Subject to some conditions, is it possible to conclude a subfield of an abelian extension generated by a unit is a cyclic extension

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    If $\rho$ is the regular representation of $G$, then $\lambda$ is the direct sum of $|G|/|H|$ copies of the regular representation of $H$. Does this not imply that if $V$ is the regular representation of $G$ modulo the trivial 1-d module, then $$\dim V^H=|G|/|H|-1?$$2012-04-15
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    Thanks! This certainly seems reasonable. I guess the bit of information I was lacking was that if $\rho$ is the regular representation of $G$ then $\lambda$ is the direct sum of $|G|/|H|$ copies of the regular representation of $H$. I guess this is pretty obvious is hindsight and I feel silly for even having asked the question. I'll need to give some thought as to how this relates to #2 at [here](http://math.stackexchange.com/questions/130951/how-can-a-subfield-of-an-abelian-extension-fail-to-be-cyclic-when-subjected-to-a). Any thoughts on this matter may be helpful.2012-04-15

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