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.