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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
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2 Answers

up vote 4 down vote accepted

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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Here is a distinct answer that is more about the group theory. It uses a small thing about simple modules over commutative rings.

Suppose G is a group in which every proper subgroup has prime power order. This means in particular, that no proper p subgroup is normalized by any q element. Assuming G is solvable, we get that Fit(G) = Oq(G) for some prime q, and that the Sylow p-subgroup P of G acts irreducibly on Fit(G). In particular, Fit(G) is an elementary abelian q-subgroup. Since P⋅Fit(G) is a subgroup not of prime power order, it must equal G, and so G is the semi-direct product of P and Fit(G). If P had a non-identity proper subgroup R, then R⋅Fit(G) would be a proper subgroup not of prime power order, so P must have order p.

Now GF(q)[P] = GF(q)[x]/(xp−1) and so Fit(G) is a quotient of this commutative ring by a maximal ideal. In particular, Fit(G) is an extension field of GF(q) generated as a vector space by the x-multiples of any non-identity element of Fit(G). Call its dimension n.

In particular, x is not contained in any proper subfield of Fit(G) and so p cannot divide qd−1 for any d < n, including d = 1 as observed by Tobias. Of course, p must divide qn−1 since it is an element of a field of order qn. These conditions are exactly that n is the multiplicative order of q mod p.

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I thought I would remark that part of my interest in these groups is mainly as a source of exercises, since quite a lot can be said about them, but they are not really important. That such groups are in fact solvable can be proven by quite elementary means (sylow plus the fact that p-groups are nilpotent), but I have so far not found an easy proof, so it might be a good (and fairly hard) exercise for an introductory course. –  Tobias Kildetoft Oct 25 '11 at 10:30
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