Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I came across the following theorem while studying representation theory over finite fields.

A cyclic group $\mathbb{Z}_n$ has a faithful irreducible representation of degree $d$ over $\mathbb{F}_p$ if and only if $n \mid p^d-1$ but $n \not\mid p^i-1$ for all $i<d$

I can show that under these conditions a faithful irreducible representation of degree $d$ over $\mathbb{F}_p$ exists by sending a generator of the cyclic group to the companion matrix of a polynomial of degree $d$ in $\mathbb{F}_p[x]$.

My question is

a) How can I show that all faithful irreducible representations of degree $d$ can be found in this way.

b) How can I show that there are no faithful irreducible representation of other degrees.

Thank you very much in advance.

share|improve this question
    
I've edited interchanging $i$ and $d$ in the divisibility conditions; since $p-1$ divides every $p^d-1$, what was there initially made non sense. –  Marc van Leeuwen Jun 1 '13 at 10:22
    
thank you, that is what I had meant to write. –  john Jun 1 '13 at 10:34
    
Here's a related question. Faithfulness is implied there by imbedding the cyclic group into the multiplicative group of an extension field, which may in turn be viewed as a group of $d\times d$ matrices over the prime field. This more or less answers the part a) of your question, I think? Look at the summands of the group algebra! –  Jyrki Lahtonen Jun 1 '13 at 17:29
    
I think that part b) also follows from that same piece of structure theory of the group algebra. The lower dimensional summands cannot give a faithful representation, because they correspond to cyclotomic cosets of size less than $d$, and those necessarily correspond to sets of non-primitive roots of unity. –  Jyrki Lahtonen Jun 1 '13 at 17:35

1 Answer 1

up vote 4 down vote accepted

A finite dimensional representation of a cyclic group $C_n$ in a $K$-vector space $V$ is determined up to isomorphism by (the conjugacy class of) the image $g\in GL(V)$ of a generator of the group. If the representation is irreducible then the minimal polynomial of $g$ must be irreducible (if it had a nontrivial proper divisor $P\in K[X]$ then $\ker(P(g))$ would be a proper sub-representation). Conversely an endomorphism$~\eta$ of a vector space$~V$ with irreducible minimal polynomial$~\mu$ has no non-trivial stable subspaces precisely when $\deg\mu=\dim V$, in other words when the minimal polynomial coincides with the characteristic polynomial (this follows from the classification of modules over the PID $K[X]$, but also elementarily because for any nonzero vector$~v$ the minimal polynomial$~P$ such that $P(\eta)(v)=0$ has$~P\mid\mu$, whence $P=\mu$ by irreducibility, and $\dim K[\eta](v)=\deg P=\dim V$, so $v$ is not containined in any proper stable subspace). The $K[X]$-module $V$ with $X$ acting as$~\eta$ is isomorphic to $K[X]/(\mu)$, with $X^i$ corresponding to $g^i(v)$ for $0\leq i<\deg\mu$; in this basis the matrix $X$ acts by the companion matrix of$~\mu$.

The minimal polynomial of the image$~g$ of a chosen generator of$~C_n$ must divide $X^n-1$. Therefore the isomorphism classes of irreducible representations of $C_n$ over$~K$ are given by the irreducible factors$~P_i$ of the polynomial $X^n-1$ in $K[X]$; the dimension of the representation is given by the degree of the factor$~P_i$. When $P_i$ does not divide $X^d-1$ for any divisor$~d$ of$~n$, which implies $P_i$ divides (the image in $K[X]$ of) the $n$-th cyclotomic polynomial, then the representation is faithful. The representation is isomorphic to the one on the field $K[X]/(P_i)$ given by multiplication by the image of$~X$, a primitive $n$-th root of unity.

All this holds over an arbitrary field$~K$. When $\def\Fp{\Bbb F_p}K=\Fp$, the irreducible polynomials of degree$~d$ (with the exception of $X$, if $d=1$) are precisely the irreducible factors of $X^{p^d-1}-1$ that do not divide $X^{p^i-1}-1$ for any $i<d$; these are the minimal polynomials of the elements of the extension field $\Bbb F_{p^d}$ that are not contained in any proper sub-field of it. The polynomial $X^n-1$ divides $X^{p^d-1}-1$ if and only if $n$ divides $p^d-1$. This does not happen at all when $p$ divides $n$; otherwise the smallest $d$ for which this happens is given by the multiplicative order of$~p$ in $(\Bbb Z/n\Bbb Z)^\times$. The irreducible faithful representation of $C_n$ over$~\Fp$ are all of that degree$~d$; there are $\phi(n)/d$ different such representations.

share|improve this answer
    
Thanks. A bit intense from my perspective but a lot of representation theory over finite or non-algebraically closed fields seems to be. Just a quick question though, what is $\phi (n)$? It doesn't seem to be defined anywhere. –  john Jun 2 '13 at 0:49
    
The name $\phi$ is the traditional one for the Euler totient function. The number $\phi(n)$ gives the number of different generators of the cyclic group $C_n$, and is also the degree of the $n$-th cyclotomic polynomial $\Phi\in\Bbb Z[X]$. –  Marc van Leeuwen Jun 2 '13 at 6:38
    
And as for the "intense" aspect, I just wanted to explain everything as elementarily as is reasonable (and to convince myself I wasn't jumping to false conclusions). A lightweight summary is that a faithful irreducible representation of a cyclic group has to diagonalise over a suitable extension of the base field, with primitive $n$-th roots of unity as eigenvalues (at the generator of the group); a single eigenspace over the minimal necessary extension field models the representation, and its dimension over the base filed equals the degree of that extension. –  Marc van Leeuwen Jun 2 '13 at 6:50

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.