# (Hypothetical) simple group of order 1004913

In 6.2.29 of Dummit and Foote, the exercise is to prove a group of order $3^3*7*13*409$ is not simple by using the permutation representation of degree 819. After having calculated the uniquely possible Sylow numbers (2863, 47853, 25767, and 819 respectively), the isomorphism type of $P_3∈Syl_3(G)$ (that is, $Z_3^3$), the list of elements by order (all prime, and there is no intersection between Sylow subgroups) and how many conjugacy classes $G$ has (1 for the identity, 2 each for the elements of order 3 [implying that all the subgroups of order 3 are conjugate] and of order 7, 4 for elements of order 13, and 136 for the elements of order 409) and how the elements manifest as permutations (order 3's as 270 3-cycles, order 7's as 117 7-cycles, order 13's as 63 13-cycles, and order 409's as 2 409-cycles), I still can't find a contradiction.

The normalizers in $S_{819}$ of all the subgroups of $G$ check out as far as Lagrange applied to their normalizers in $G$ is concerned, and the Sylow 409-subgroups of $S_{819}$ and their normalizers don't appear to yield further results. What's missing?

• An element $g$ of order 409 must fix one point, say $a$, and have two orbits of length 409. Let $h$ be an element of order 3 that normalizes $\langle g \rangle$. Then $h$ must fix $a$ and exactly one point in each of the two long cycles of $g$, which makes 3 points altogether, contradicting your assertion that it has 270 3-cycles (and hence 9 fixed points). – Derek Holt Apr 18 '13 at 20:54
• Is that correct? I used another method to obtain my result; since each Sylow 3-subgroup contains 13 subgroups of order 3, and none of these intersect, there are 37219 subgroups of order 3. Since $N_G(P_{409})$ of order $3*409$ is nonabelian, not taking intersections into account there $409*819=334971$ groups of order 3, meaning that there must be a subgroup of order 3 contained in 9 normalizers or more; since these 3-subgroups are conjugate, they are all in precisely 9, and thus their representation by action on $Syl_{409}(G)$ should provide 9 fixed indices, right? – Feryll Apr 18 '13 at 21:00
• Seems OK, so that seems to be a contradiction! – Derek Holt Apr 18 '13 at 22:11
• Very well, I can understand it in this way. Though, I had trouble justifying your statement at first, is this the easiest way to think about it?: Assume $h$ fixes two or more points in a long cycle of $g$ (in fact it must fix at least one in one of them, because it acts as 3-cycles on the 819 points, it fixes $a$, and $3 \not \mid 818$), and call them $b$ and $c$. But we have $hc=hg^{n_1}b=h(h^{-1}g^{n_2}h)b=g^{n_2}b≠g^{n_1}b=c$, a contradiction. Or is there a generalization I'm missing? – Feryll Apr 18 '13 at 22:40
• This was discussed a long time ago on the Ask an Algebraist board. Using Derek's hint, one can show every element of order $3$ fixes exactly 3 points. The Cauchy-Frobenius theorem then yields a non-integer number of orbits, which is a contradiction. The AAT link is: at.yorku.ca/cgi-bin/… – user641 Apr 19 '13 at 2:45