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Question. Which groups of order $100$ are Frobenius groups?

The OEIS says that there are two Frobenius groups of order $100$, but I am finding three of them and I'd be grateful if someone can point out where I am going wrong.

(I was not able to locate an independent source for the Frobenius groups of order $100$, and my results agree with the OEIS up to order $99$.)

The three groups that I am finding are Frobenius are those with small group IDs [[100, 3], [100, 11], [100, 12]].

In each of these cases, I get the following:

  • the Frobenius kernel $K$ is the (necessarily abelian) Sylow $5$-subgroup, which is also the Fitting subgroup. This is (uniquely, up to conjugacy) complemented, by Schur-Zassenhaus. (For SmallGroup(100,3), $K\simeq C_{25}$, for the other two $K\simeq C_5^2$.)
  • the Frobenius complement $H$ is the Sylow $2$-subgroup of order $4$
  • in all three cases, $H$ is core-free
  • in all three cases, the action of the group on $H$ is Frobenius

The first three items are easy to verify. For the last, these are the permutation representations that I get on the cosets of the putative complement $H$.

For SmallGroup( 100, 3 ):

$$\langle (2,8,25,21)(3,6,15,10)(4,20,24,9)(5,18,23,12)(7,19,22,11)(13,16,17,14), (2,25)(3,15)(4,24)(5,23)(6,10)(7,22)(8,21)(9,20) (11,19)(12,18)(13,17)(14,16)\rangle$$

For SmallGroup( 100, 11 ): $$\langle (2,4,11,7)(3,6,15,10)(5,13,25,21)(8,20,24,14)(9,22,23,12)(16,17,19,18), (2,11)(3,15)(4,7)(5,25)(6,10)(8,24)(9,23)(12,22)(13, 21)(14,20)(16,19)(17,18), (1,2,4,7,11)(3,5,8,12,16)(6,9,13,17,20)(10,14,18,21,23)(15,19,22,24,25), (1,3,6,10,15)(2,5,9,14,19)(4,8,13,18,22)(7,12,17,21,24)(11,16,20,23,25) \rangle$$

For SmallGroup( 100, 12 ):

$$\langle (2,7,11,4)(3,6,15,10)(5,17,25,18)(8,9,24,23)(12,20,22,14)(13,19,21,16), (2,11)(3,15)(4,7)(5,25)(6,10)(8,24)(9,23)(12,22)(13, 21)(14,20)(16,19)(17,18), (1,2,4,7,11)(3,5,8,12,16)(6,9,13,17,20)(10,14,18,21,23)(15,19,22,24,25), (1,3,6,10,15)(2,5,9,14,19)(4,8,13,18,22)(7,12,17,21,24)(11,16,20,23,25)\rangle$$

In each case, I check that the group is transitive, that the stabiliser of $1$ is non-trivial (as it must be), and that the two-point stabilisers are all trivial.

I've also checked (redundantly) that $C_G(h)\leq H$, for all $h\in H\setminus 1$, and that $C_G(k)\leq K$, for all $k\in K\setminus 1$, in each case.

I have replicated these calculations in two computer algebra systems with the same results, so I suspect that, rather than a software bug, there is a bug somewhere in my understanding.

James
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    Yes I agree with your results - there are three Frobenius groups of order $100$ (up to group isomorphism). – Derek Holt Sep 30 '18 at 12:15
  • @DerekHolt Thanks for looking at this. Is there some way you can turn this into an answer so I can accept it? – James Oct 01 '18 at 16:43

2 Answers2

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I agree with your calculation. A routine search through the small groups library shows that only the groups $\mathtt{SmallGroup}(100,i)$ for $i=3,11,12$ are Frobenius groups.

Or you could prove it theoretically. It is easy to see that the only possible order of a Frobenius kernel $K$ in such a group is $25$. A Frobenius complement must be cyclic of order $4$ and act fixed point freely by conjugation on $K$.

If $K$ is cyclic then ${\rm Aut}(K)$ is cyclic and there is only one possible action.

Otherwise $K$ is elementary abelian and ${\rm Aut}(K) \cong {\rm GL}(2,5)$. Elements of order $4$ are diagonalizable, and there are two two conjugacy classes of subgroups of order $4$ with fixed point free action, with representatives $\left\langle \left(\begin{array}{cc}2&0\\0&2\end{array} \right)\right\rangle$ and $\left\langle \left(\begin{array}{cc}2&0\\0&3\end{array} \right)\right\rangle.$

Derek Holt
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There are three non-isomorphic Frobenius groups of order 100, two with kernel K = $C_{5}^2$. Let h be an element of order 4 in FG of order 100 with kernel K. If h is in the FG of order 600 with cyclic complement, the conjugate by h of every k in K is $k^2$ (or, isomorphically, $k^{-2}$). If h is in the FG of order 600 whose complement has normal subgroup Q8, we can label the six subgroups of order 5 as A, B, C, D, E, F use $k_A$ for a generator of A, k*h as the conjugate of k by H, so that $k_A$*h = $k_A^2$, $k_B$*h = $k_B^{-2}$, $k_C$*h = $k_D$, $k_D$*h = $k_C^{-1}$, $k_E$*h = $k_F$, $k_F$*h = $k_E^{-1}$.

Ken Lebensold
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  • Perhaps the downvoters could provide some feedback? This reads like a serious attempt at a non-trivial problem, and which backs up a claim on the OEIS (so adheres to the status quo). So this answer is surely worth a pause for thought by those in the know, rather than a quick downvote without comment? – user1729 Apr 21 '20 at 16:26
  • I would imagine that the downvotes are for suggesting that there could be error in the small groups database! But what the poster is saying here is correct in that all groups that arise in the way described are isomorphic: in fact this is $\mathtt {SmallGroup}(100,12)$, which is the second of the two groups that I described in my answer to the question. But this is not isomorphic to $\mathtt {SmallGroup}(100,11)$, which is the first of the two groups. It is not hard to see that they are not isomorphic, because the action of the complement is scalar in one case and nonscalar in the other. – Derek Holt Apr 21 '20 at 18:42
  • Thank you for these comments. I've edited out any reference to the SmallGroup listing, which I know nothing about and should not have included in the first place. – Ken Lebensold Apr 22 '20 at 13:32
  • My comment was based on an assumption that there is only one Frobenius group of order 600, 2-Sylow subgroup Q8, and a group of order 100 with the same kernel would be a subgroup of that. I have since learned that there is also a group of order 600 with cyclic complement, and although I haven't checked it out in detail, its subgroup of order 100 is likely not the same as that other group, in which case James and Derek Holt are right that there are three non-isomorphic Frobenius groups of this order. – Ken Lebensold May 15 '20 at 14:13