Search for Simplicity (6.2.16)

Dummit and Foote Abstract Algebra, section 6.2, exercise 16:

Prove there are no simple groups of odd composite order < 10,000.

Proof: I will record some difficult cases:

1575: Sylow analysis gives us $n_5=21$. Since $21≠1 \mod{5^2}$, we have $P_5∩Q_5$ of order 5 and due to index crunching we obtain $|N_G(P_5∩Q_5)|=3*5^2$. But now $n_5(N_G(P_5∩Q_5))=1$, even though $P_5≠Q_5$ and $P_5,Q_5∈Syl_5(N_G(P_5∩Q_5))$.

3465: Lemma 1: Let $G$ be a group, let $P∈Syl_p(G)$ and assume $N_G(P)$ is cyclic. Then there are precisely $n_p \cdot φ(|N_G(P)|)$ elements of order $|N_G(P)|$ in $G$. Proof: Since $P \trianglelefteq N_G(P)$, we have distinct normalizers of Sylow p-subgroups for distinct Sylow p-subgroups, so that there are $n_p$ distinct normalizers. As well, if $|x| = |N_G(P)|$, then for some $Q∈Syl_p(G)$ we have $Q \trianglelefteq \langle~x~\rangle$ so that $\langle~x~\rangle=N_G(Q)$, and now every element of such an order is contained in a normalizer. Since these normalizers are conjugate to one another and are thus cyclic as well, we have each normalizer contains $φ(|N_G(P)|)$ elements of the specified order, and since no two distinct normalizers share elements of this order, we have the lemma proven.$~\square$

Lemma 2: A group $G$ of order $231=3*7*11$ has an element of order 33. Proof: We have $P_{11} \trianglelefteq G$, so that $P_3P_{11} ≅ Z_{33}$ is a subgroup of $G$.$~\square$

By Sylow analysis on 3465, we obtain:$$n_3∈\{7,55,385\}~~~~~n_5∈\{11,21,231\}~~~~~n_7∈\{15,99\}~~~~~n_{11}=45$$This reveals that $N_G(P_{11}) ≅ Z_{77}$ (in particular, this implies $G$ has no elements of order 33). By the first lemma, there are $45*60=2700$ elements of order 77 in $G$. By $n_{11}=45$, we obtain 450 elements of order 11. Now, if $n_7=99$, this would produce 594 elements of order 7, which overloads the order of $G$. Therefore $n_7=15$, and now $|N_G(P_7)|=231$ and contains an element of order 33 by the second lemma, a terminating contradiction.

9765: By Sylow analysis and index crunching, we obtain:$$n_3∈\{31,217\}~~~~~n_5∈\{31,651\}~~~~~n_7=155~~~~~n_{31}=63$$We have $|N_G(P_7)|=3^2*7$. Now, if $n_5=31$, we have $|N_G(P_5)|=3^2*5*7$, so that $P_7 \trianglelefteq P_5P_7$ and then $35 \mid |N_G(P_7)|=63$, a contradiction. So $n_5=651$ and $N_G(P_5)≅Z_{15}$, so that by lemma 1 we have 5208 elements of order 15. With Sylow, we obtain at least $5208+651*4+155*6+63*30=10,632$ elements in $G$, an impossibility.$~\square$