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Let $V$ be a $n$-dimensional $\mathbb{C}$-vector space and let $G$ be a group. Suppose we have a representation $\phi: G \to \text{Aut}(V)$, then this representation makes $V$ into a $G$-module by $g.v := \phi(g)v$. My question is, what is the dimension of this $G$-module? I have a feeling that it is $n$ since $V$ is an $n$-dimensional vector space. However I am not sure and I can't find a proof for this.

Kxxxhk
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    Yes, the dimension of a $G$-module is the dimension of the vector space on which it acts. There is nothing to prove, it is just the definition. – Tobias Kildetoft Mar 02 '16 at 09:19
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    There isn't a great definition of dimension for modules over a ring that aren't free. But if you mean something like "minimal number of generators as a $\mathbb{C}[G]$ module," this number will usually be smaller than $n$. – hunter Mar 02 '16 at 09:20
  • @Tobias that isn't quite right. For example, if $\phi$ is an irreducible representation, then $V$ is at most a one-dimensional module. – Ben Grossmann Mar 02 '16 at 11:29
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    @Omnomnomnom That is only if $G$ is abelian, and how is it relevant anyway? – Tobias Kildetoft Mar 02 '16 at 11:30
  • @Kxxxhk for a module to have a "dimension" in the linear algebra sense (i.e. $V$ has a basis), you really need the module to be a free module. In particular, it should be the case that for any non-zero $v\in V$,constants $a_i$ and $g_i\in G$, we should not have $$ \left( \sum a_i \phi(g_i)\right) v = 0 $$ – Ben Grossmann Mar 02 '16 at 11:35
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    @Omnomnomnom It already has that. It is a complex vectorspace. – Tobias Kildetoft Mar 02 '16 at 11:36
  • @Tobias it is relevant because we're looking for the dimension of the module, and now we know that it generally doesn't match the dimension of the underlying space, which was your initial claim. – Ben Grossmann Mar 02 '16 at 11:37
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    @Omnomnomnom When speaking of $G$-modules, we practically always mean the dimension of the underlying vectorspace. I have yet to encounter anywhere someone caring about the dimension as a module over the group ring (since it will practically never be free). – Tobias Kildetoft Mar 02 '16 at 11:39
  • @Tobias okay! I guess it's a terminology thing, then. That makes sense now. – Ben Grossmann Mar 02 '16 at 11:40
  • @TobiasKildetoft I am a little confused here. For example, a $G$-module is said to be irreducible/simple if it doesn't have a nontrivial proper submodule. In my textbook, it states that every one-dimensional $G$-modules are irreducible. Tobias, you state that we are considering the dimension of the vector space, if so, why would the module be simple if it is one-dimensional as a vector space? I don't see this. – Kxxxhk Mar 02 '16 at 14:14
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    Because any submodule will in particular be a subspace. So it would be either the entire thing or $0$. – Tobias Kildetoft Mar 02 '16 at 14:16
  • @TobiasKildetoft So, is it correct to say that a module is irreducible if and only if it doesn't have any proper subspaces as a vector space? – Kxxxhk Mar 02 '16 at 14:17
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    No, that would be the same as being $1$-dimensional. submodules are subspaces, but not all subspaces are submodules. – Tobias Kildetoft Mar 02 '16 at 14:19
  • @TobiasKildetoft ok thanks! How can i see that every submodule is a subspace? Is the map $G \to \text{Aut}(V)$ surjective? – Kxxxhk Mar 02 '16 at 15:10
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    Not necessarily. Every submodule is by definition a subspace which is also stable under the action of the group. – Tobias Kildetoft Mar 02 '16 at 16:33
  • @TobiasKildetoft what do you mean? a submodule is by definition closed under multiplication of the group $G$, but why is it closed under its set of scalars? – Kxxxhk Mar 02 '16 at 16:44
  • Because that is part of the definition. – Tobias Kildetoft Mar 02 '16 at 17:18
  • @TobiasKildetoft No its not. If $W$ is a submodule then for every $g \in G$ we sure have that $\phi(g)v \in V$ where both $\phi(g)$ and $v$ is thought of as a $1x1$ matrix. But for instance if we have the basis $v_1,...,v_n$, in order to have $a_1v_1+...+a_nv_n$ i must have that $\phi(g)=(a_1,...,a_n)$ for some $g$. This is why I asked for surjevtivity, what am I missing? – Kxxxhk Mar 02 '16 at 19:33
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    You are missing the full definition of what a submodule is, look it up again. – Tobias Kildetoft Mar 02 '16 at 19:35

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