What are the Frobenius groups of order $100$?

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 $\left\langle \left(\begin{array}{cc}2&0\\0&3\end{array} \right)\right\rangle.$

Tags:

Group Theory

Frobenius Groups

Related

When does $\exists x \forall y \phi(x,y) \leftrightarrow \forall y \exists x \phi(x,y)$ hold? Are the real numbers the unique Dedekind-complete ordered set? A Matrices Problem - Cayley-Hamilton or Bash? Are real and complex analysis related in any way? Does $3-4+1$ equal $0$ or $-2$.. or maybe $2$? How to know that $2n^3+9n^2+13n+6$ factors into $(n+1)(n+2)(2n+3)$? Does swapping columns of a matrix cause the rows of the inverse matrix to be swapped? "Bad" Fourier Series derivation Proving that $\sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{n^{2}}=\frac{\pi^{2}}{12}$ Calculate $\sum_{n=0}^\infty \frac1{(4n^2 - 1)^2}$ Show that $\int_0^{\infty}\frac{\operatorname{Li}_s(-x)}{x^{\alpha+1}}\mathrm dx=-\frac1{\alpha^s}\frac{\pi}{\sin(\pi \alpha)}$ Does the elliptic curve $y^2 = 4 x^3 -6075$ have any integer points?

Recent Posts