# Irreducible representations of a cyclic group of order p over a field of q elements when p and q are distinct primes

What is a sufficient condition for the existence of an irreducible representation of degree $n$ of the cyclic group of order $p$ over the field of $q$ elements when $p$ and $q$ are distinct primes?

Some obvious necessary conditions (for a non-trivial representation) are that $p$ must divide $\frac{q^n-1}{q-1}$ and $n \leq p$, but I have not been able to come up with a sufficient condition (apart from $n=1$).

The motivation is that such a representation gives rise (via the corresponding semidirect product) to a group, whose order is not the power of a prime, but such that the order of any proper subgroup is the power of a prime, and all such groups arise this way.

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Another way to phrase the exact condition is that p must divide the nth cyclotomic polynomial evaluated at q. Not only do you divide by q-1, but also by all the smaller ones: q+1, qq+1, qq+q+1, qq-q+1, etc. that divide q^n-1. So if n = 4, then p must divide qq+1; if n = 2, then p must divide q+1; if n=6, then p must divide qq-q+1. –  Jack Schmidt Oct 24 '11 at 15:26

A necessary and sufficient condition for the cyclic group of order p to have a faithful irreducible representation of dimension n over a field with q elements (where p is prime, q is a prime power, and n is a positive integer) is:

n is the multiplicative order of q mod p.

Equivalently,

p divides the value of the nth cyclotomic polynomial evaluated at q.

The cyclic group of order p has exactly p absolutely irreducible representations over an algebraically closed field of characteristic dividing q (assuming that p, q are relatively prime). Each is of dimension one, and is parameterized by the eigenvalue of a generator of the cyclic group: a pth root of unity inside the algebraic closure of the field with q elements.

If the multiplicative order of q mod p is n, then the non-identity eigenvalues generate a field of order qn over the field of q elements. Writing such eigenvalues as matrices over the field of q elements gives an irreducible (but only absolutely irreducible when n = 1) representation of dimension n.

For instance, when p is 7, then we have the following cases:

• 0 ≡ q mod 7: one absolutely irreducible representation of degree 1
• 1 ≡ q mod 7: seven absolutely irreducible representations of degree 1
• 2 ≡ q mod 7: one absolutely irreducible representation of degree 1, two irreducible representations of degree 3
• 3 ≡ q mod 7: one absolutely irreducible representation of degree 1, one irreducible representation of degree 6
• 4 ≡ q mod 7: one absolutely irreducible representation of degree 1, two irreducible representations of degree 3
• 5 ≡ q mod 7: one absolutely irreducible representation of degree 1, one irreducible representation of degree 6
• 6 ≡ q mod 7: one absolutely irreducible representation of degree 1, three irreducible representations of degree 2

We can decompose the regular representation over the algebraic closure: it is just diagonal matrices with p distinct pth roots of unity as entries. Over the field with q elements, we'd need the matrix entries to be stable under qth powers.

For instance, if 6 ≡ q mod 7, then we get: $$\begin{bmatrix} \zeta_7^0 &.&.&.&.&.&.\\ .&\zeta_7^1&.&.&.&.&.\\ .&.&\zeta_7^{-1}&.&.&.&. \\ .&.&.&\zeta_7^2&.&.&. \\ .&.&.&.&\zeta_7^{-2}&.&. \\ .&.&.&.&.&\zeta_7^3&. \\ .&.&.&.&.&.&\zeta_7^{-3} \end{bmatrix}$$ is conjugate to $$\left[\begin{array}{r|rr|rr|rr} 1 &.&.&.&.&.&.\\ \hline .&.&1&.&.&.&.\\ .&\zeta_7^1 + \zeta_7^{-1}&-1&.&.&.&. \\ \hline .&.&.&.&1&.&. \\ .&.&.&\zeta_7^2+\zeta_7^{-2}&-1&.&. \\ \hline .&.&.&.&.&.&1 \\ .&.&.&.&.&\zeta_7^3+\zeta_7^{-3}&-1 \end{array}\right]$$ and this matrix is unchanged by replacing its entries with their qth powers (which is a field automorphism in fields whose characteristic divides q). Note how the eigenvalues get paired up, since 6 ≡ q mod 7 has multiplicative order 2.

For 2 ≡ q mod 7, they would get matched up in triples:

$$\begin{bmatrix} \zeta_7^0 &.&.&.&.&.&.\\ .&\zeta_7^1&.&.&.&.&.\\ .&.&\zeta_7^{2}&.&.&.&. \\ .&.&.&\zeta_7^4&.&.&. \\ .&.&.&.&\zeta_7^{-1}&.&. \\ .&.&.&.&.&\zeta_7^{-2}&. \\ .&.&.&.&.&.&\zeta_7^{-4} \end{bmatrix}$$ is conjugate to $$\left[\begin{array}{r|rrr|rrr} 1 &.&.&.&.&.&.\\ \hline .&.&1&.&.&.&.\\ .&.&.&1&.&.&. \\ .&\omega&\omega^*&-1&.&.&. \\ \hline .&.&.&.&.&1&. \\ .&.&.&.&.&.&1 \\ .&.&.&.&\omega^*&\omega&-1 \end{array}\right]$$ where $\omega = \zeta_7^1 + \zeta_7^{2}+\zeta_7^4$ and $\omega^* = \zeta_7^{-1} + \zeta_7^{-2}+\zeta_7^{-4}$ are exchanged by "complex conjugation" but are fixed under the qth Frobenius map.

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