What are the Frobenius groups of order $100$? 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.
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 $\left\langle \left(\begin{array}{cc}2&amp;0\\0&amp;2\end{array} \right)\right\rangle$ and $\left\langle \left(\begin{array}{cc}2&amp;0\\0&amp;3\end{array} \right)\right\rangle.$
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 = $k_E^{-1}$.