A question about dimension and irreducibility of a representation Let $H$ be a non-trivial group of order $m$, and $\phi:H\to GL(V)$ a complex representation of dimension $m$. How to prove that $\phi$ is not irreducible?
Here is what I thought: If $V$ is a $\mathbb{C}H$-module, generated by $a_1,...,a_m$, then it's a vector space generated by $\{ ga_i| g\in G\}$, now since $|H|=m$ and $V$ as a $\mathbb{C}$-vector space is of dimension $m$, we see $V$ is a $\mathbb{C}H$-module with one generator, i.e. $V$ is a cyclic module. But then it suffices to show every cyclic $\mathbb{C}H$-module is not irreducible? I wonder if it's true.
 A: Because $V$ is irreducible, $V$ will be a isomorphic to a subrepresentation of $k[G]$, functions on $G$, with right translation of argument (true for any field $k$, tedious details below). Assume $V$ has dimension $|G|$. We conclude $V$ is isomorphic to  $k[G]$, so $k[G]$ irreducible. We have the trivial representation inside $k[G]$. That implies $\dim k[G]=1$, so $|G|=1$.

 Indeed, let  $l$ be  a linear functional on $V$. Consider the map
$$v\mapsto \phi_v(\cdot), \ \ \phi(h)= l(h v)$$
 One checks that 
  $$\phi_{gv}(h)= l(h(gv))=l(hg v) = \phi_{v}(hg)= g\cdot (\phi_v)(h)$$
  so $\phi$ is a morphism of representations. Since $V$ is irreducible and
  $l\ne 0$, the map is injective.   

A: Here is the most straightforward way to show it that I know of. 
Take $v\in V\backslash\{0\}$. Then (as you described), $V = <\{gv:g\in H\}>$. Since $\dim(v)=|H|\geq |\{gv:g\in H\}|$, $\{gv:g\in H\}$ is a basis for $V$ (As a remark, this is isomorphic to the regular representation). But then $V$ has a non-trivial proper submodule $<\displaystyle\sum_{g\in H}gv>$.
Note: All irreducible modules are cyclic, so you cannot show what you suggested.
