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The post Irreducible representations of a cyclic group over a field of prime order discusses the irreducible representations of a cyclic group of order $N$ over a finite field $\mathbb{F}_p$ where $N$ does not divide $p$.

Where can I find information about the irreducible representations in the case where $p$ does divide $N$?

(I'm interested in this because I'm wondering if, given a finite dimensional vector space $\mathbb{F}_{p}^{M}$ if there are examples of particularly simple invertible linear operators $T$ such that $T$ does not preserve a subspace. Since the vector space contains finitely many elements it seems that this is the same thing as an irreducible representation of a cyclic group.

However, based on the post I read above, where p does not divide N, it seems like T is just multiplication by a primitive element that generates the extension $\mathbb{F}_{q^{M}}$.)

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I don't understand what your question has to do with your motivation. You have already found the thing you say you are looking for. –  Qiaochu Yuan Jul 7 '12 at 20:51
    
Judging from the last paragraph you seem to be well on your way to having solved your problem. That's indeed what happens. With the proviso that the order of $T$ may also have $p$ as a factor, which you can get rid of by using $T^p$ instead. If $T$ is of order (a power of) $p$, then $1$ is its sole eigenvalue, and it will stabilize any subspace of the eigenspace. –  Jyrki Lahtonen Jul 7 '12 at 20:59
    
I have a vague recollection that Curtis & Reiner cover this briefly. –  Jyrki Lahtonen Jul 7 '12 at 21:00

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up vote 2 down vote accepted

The question of how to classify all finite-dimensional representations of $C_n$ over an arbitrary field $F$ can be studied using the structure theorem for finitely-generated modules over a principal ideal domain, in this case $F[x]$. The structure theorem asserts that any finitely-generated module is uniquely a finite direct sum of modules of the form $F[x]/p(x)^r$ where $p \in F[x]$ is irreducible and $r$ is a non-negative integer.

If $T$ is an operator acting on $F^k$ for some $n$, then $F^k$ becomes a finitely-generated module over $F[x]$ with $x$ acting by $T$. $T$ gives a representation of the cyclic group $C_n$ if and only if $T^n = 1$, in which case the summands $F[x]/q(x)^r$ in the decomposition of $F^k$ must have the property that $q(x)^r | x^n - 1$.

If $F$ has characteristic $0$ or has characteristic $p$ and $p \nmid n$, then $x^n - 1$ is separable over $F$, hence $r \le 1$ and $F^k$ is a direct sum of irreducible representations, all of which are of the form $F[T]/q(T)$ where $q$ is an irreducible factor of $x^n - 1$ over $F$.

If $F$ has characteristic $p$ and $p | n$, then writing $n = p^s m$ where $p \nmid m$ we have $$x^n - 1 = (x^m - 1)^{p^s}$$

from which it follows that $r \le p^s$ (but now it is possible to have $r > 1$). If $r > 1$, then the corresponding representation $F[T]/q(T)^r$ is indecomposable and not irreducible, where $q$ is an irreducible factor of $x^m - 1$ over $F$. The irreducible representations occur precisely when $r = 1$. In other words,

The irreducible representations of $C_{p^s m}$, where $p \nmid m$, over a field of characteristic $p$ all factor through the quotient $C_{p^s m} \to C_m$.

One can also see this more directly as follows. If $V$ is an irreducible representation of $C_{p^s m}$ over a field of characteristic $p$ and $T : V \to V$ is the action of a generator, then $$T^{p^s m} - 1 = (T^m - 1)^{p^s} = 0.$$

Thus $T^m - 1$ is an intertwining operator which is not invertible, so by Schur's lemma it is equal to zero.

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In general, if a finite group $G$ has a normal subgroup $U$ whose order is power of a prime $p,$ and $G$ acts irreducibly on finite dimensional vector space $V$ over a field $F$ of characteristic $p,$ then $U$ has a non-zero fixed vector on $V,$ while on the other hand, the fixed-point subspace $V^{U}$ of $U$ on $V$ is easily checked to be $G$-invariant, so we must have $V^{U} = V$ by irreducibility as $V^{U}$ is non-zero. This is a well-known fact, and a more general version of the remark at the end of Qiaochu's answer. The argument when $V$ has finite cardinality is easier considering orbit lengths on non-zero vectors. However, in general, to prove that $V^{U} \neq \{0 \}$ (without further assumption) we can proceed by induction on $|U|.$ The case that $U$ has order $p$ is just linear algebra, as a generator of $U$ has the eigenvalue $1$. If $|U| >p,$ let $Z$ be a subgroup of $Z(U)$ of order $p,$ which is possible as $Z(U) \neq 1$ (since $U$ is $p$-group). If $Z \neq U,$ then we still have $Z \lhd U$ as $Z$ is characteristic in $U.$ By induction $V^{Z} = V,$ and then we can replace $G$ by $G/U$ and $U$ by $U/Z$ to obtain the desired inclusion by induction.

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Thanks for the information. I have a couple of questions: 1) How does the linear algebra work in the base case (as it is not over an algebraically closed field like in our introductory classes)? I assume you are saying if $a$ generates the cyclic group $C_{p}$, then $a$ has an eigenvalue of 1 because $I-a$ is nilpotent? How do you show this? 2) Instead of induction, would it be sufficient to take an arbitrary vector $v$ in $V$ and then take the vector space generated over ${F}_{p}$ by the orbit of $v$ under the action of $U$ and use the orbit lengths mentioned earlier? –  Dtseng Jul 8 '12 at 1:11
    
For 1) do we use $(I-a)^p=I-a^{p}=0$? –  Dtseng Jul 8 '12 at 1:18
    
Yes, you do need an algebraically closed field, because $1$ is an eigenvalue, an $1$ is every field. Your second questin gives the answer to the first question of your first. As for the second part, I think your argument is OK, you can replace $V$ by a finite vector space to prove the existence of a fixed vector of $U.$ –  Geoff Robinson Jul 8 '12 at 7:50

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