# Non-isomorphic representations of equal degree (namely, of $C_4$)

I'm learning about Representation Theory, and I've come across a statement I don't understand:

"If $\rho$ and $\rho'$ are isomorphic representations, then they have the same dimension. However, the converse is not true: in $C_4$, there are four non-isomorphic 1-dimensional representations. If $\omega = e^{2 i \pi /4}$, then we have $\rho_j(\omega^i) = \omega^{ij} \ (0 \leq i \leq 3)$"

I do not understand how the $p_j$ are representations of $C_4$; as I understood it, a representation of a group was a homomorphism from the group into a group of linear transformations of vector spaces (or at least into the automorphism group of some object). But these are homomorphisms from $C_4$ into some subgroup of $C_4$. Any explanation would be appreciated.

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You do notice that $\omega$ is a complex number. A complex number can be identified with a 1x1 matrix. Granted, this was unnecessarily confusing. I would use a different notation for the generator of $C_4$, say $g$, when we would have $\rho_j(g^i)=\omega^{ij}$. So I am siding with you to that extent! –  Jyrki Lahtonen Jan 17 '12 at 20:12
The map in that statement is $\rho_j:C_4\to\mathrm{GL}(1,\mathbb C)$. Now $\mathrm{GL}(1,\mathbb C)$ is canonically isomorphic to the multiplicative group $\mathbb C^\times$, so to give $\rho_j$ we may just as well given an homomorphism $\rho_j:C_4\to\mathbb C^\times$.
You seem to have defined $C_4$ as the group of $4$th roots of unity, which is a subgroup of $\mathbb C^\times$, and then it is easy tosee that the image of a group homomorphism $C_4\to\mathbb C^\times$ will always have —as you noticed— its image contained in $C_4$.
This makes perfect sense. If only "$\mathrm{GL}(1,\mathbb C)$" had appeared somewhere in the sentence... (it's on the 4th page of introductory notes on Representation Theory; I would have expected a bit more clarity!) –  Matt Jan 17 '12 at 20:24