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I just wonder any sources that provides the list of all minimal irreducible subgroups of $\text{GL}(2,p)$. Here the term "minimal" means there is no proper subgroup that is irreducible. A list of maximal subgroups of $\text{PSL}(2,p)$ is available here. For some case, it is easy to know the subgroup is reducible, but generally I am not sure how to dig information from here. I think it would be good if there is some reference which provides full list of all irreducible subgroups.

I think first we need to know all the maximal subgroups of $\text{GL}(2,q)$. Since $\text{GL}(2,q)'=\text{SL}(2,q)$, so a maximal subgroup $M$ of $\text{GL}(2,q)$ contains $Z$ or $\text{SL}(2,q)$. If $M>Z$, then we look at the subgroups of $\text{PGL}(2,q)$; if $M>\text{SL}(2,q)$, then we look at the subgroups of $\text{PSL}(2,q)$. Then the next thing is that if you know a group is irreducible, how do you analyse the irreducibility of its subgroups? Since sometimes it seems not clear to me that how the subgroup acts on the vectors space. And I am also not too sure how to find the critical subgroups, that is the minimal one. Do we have any trick for treating this?

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What do you mean when you say an "irreducible subgroup"? – Alexander Gruber Mar 14 '13 at 7:49
@Alexander, regarding $\text{GL}(2,p)$ acts on the $2$-dimensional vector space $\text{GF}(p)^2$. The subgroup that is irreducible if it doesn't fix any proper subspace. – Easy Mar 14 '13 at 8:14

The subgroups of ${\rm PSL}(2,q)$ for prime powers $q$ were classified by Dickson in 1901. You can find the lists and proofs in Huppert's German book "Endliche Gruppen I". Of course you need the subgroups of ${\rm PGL}(2,p)$, which is twice as big, and then you need to check their inverse images in ${\rm GL}(2,p)$ for irreducibility. So you have some work to do!

I will have a go at listing the minimal irreducible groups, but it is a little complicated, and I may not get it absolutely right.

  1. For odd primes $r$ dividing $p+1$, the cyclic group of order $r$.

  2. The cyclic group of order $2^r$, where $2^{r-1}$ is the highest power of 2 dividing $p-1$.

  3. The dihedral group of order $2r$ for odd primes $r$ dividing $p-1$.

I believe that is all when $p \equiv 3 \pmod 4$. (That case is simpler, because all elements of order 4 act irreducibly.) It is more complicated when $p \equiv 1 \pmod 4$. In that case, you also get:

  1. The dihedral group $D_8$.

  2. The quaternion group $Q_8$.

  3. For odd primes $r$ dividing $p-1$, a group of order $4r$ contained in ${\rm SL}(2,p)$.

  4. This is the complicated one. After doing some computations with $p=17$, I think, for all powers $2^r$ dividing $p-1$ with $r \ge 3$, there is a nonabelian group of order $2^{r+1}$ intersecting the scalar subgroup with order $2^{r-1}$.

Here is a bit more information on methods used. The maximal subgroups of ${\rm GL}(2,p)$ (with $p$ prime) are the imprimitive groups of order $2(p-1)^2$, the semilinear groups of order $2(p^2-1)$, groups whose intersection with ${\rm SL}(2,p)$ is a double cover of $A_4$, $S_4$ or $A_5$, and groups containing ${\rm SL}(2,p)$. Since ${\rm SL}(2,p)$ itself is never minimal irreducible, we can forget about that, and the ones involving $A_4$, $S_4$ or $A_5$ all have a normal irredcuible subgroup $Q_8$, which is itself imprimitive and/or semilinear, so we can effectively forget about those.

So we can restrict attention to imprimitive and semilinear subgroups. The maximal semilinears have a cyclic subgroup of order $q^2-1$ which acts irreducibly but not absolutely irreducibly, and it is not hard to see that their minimal irreducible subgroups are those in classes 1 and 2 above.

So it remains to find the minimal irreducible subgroups of the maximal imprimitive subgroups. These are all conjugate, and one of them, $H$ say, consists of the monomial matrices, so we can just look for subgroups of $H$. So the group $D$ of diagonal matrices has index 2 in $H$ and is reducible. So a minimal irreducible $K$ will have a subgroup of $D$ of index 2. If $|K \cap D|$ is odd, then get the groups in class 3 above. As I noted before, if $p \equiv 3 \pmod 4$, then elements of order 4 act irreducible, and any other such subgroup would have an element of order 4, so would not be minimal irreducible. But it is more complicated when $p \equiv 1 \pmod 4$.

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good effort! Before we proceed further, I just generally wonder what kind of technique you used? See my additional elaboration in the post. Thanks – Easy Mar 15 '13 at 11:51
I have added a bit more information on this, which I hope is helpful. – Derek Holt Mar 15 '13 at 15:10

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