2

Before this writing the smallest order for which the nonexistence of simple groups of that order is not explicitly demonstrated on this site is 480; this self-answered question aims to fill that gap. (See Prove there are no simple groups of even order $<500$ except orders $2$, $60$, $168$, and $360$., the sole answer of which describes an outline for proving the titular statement but does not describe how to establish the claim for order $480$. In particular, this question-answer pair completes that outline in the sense that all cases not explicitly resolved therein are covered by other, linked answers on this site.) Other answers are, of course, welcome.

How does one show that there are no simple groups of order $480$?

Travis Willse
  • 108,056

1 Answers1

5

Suppose $G$ is a simple group of order $480 = 2^5 \cdot 3 \cdot 5$.

Since $G$ is simple, if it has a proper subgroup of index $k$, then $G$ embeds into $S_k$; as $480 \not\mid |S_k|$ for $k \leq 7$, any proper subgroup of $G$ has index $\geq 8$.$^*$ Sylow's Third Theorem then gives that the count $n_2 := |\operatorname{Syl}_2(G)|$ satisfies $n_2 \mid 15$ and $n_2 = |P : N_G(P)| \geq 8$ (for any, equivalently, every $P \in \operatorname{Syl}_2(G)$), leaving only the possibility$$n_2 = 15 .$$

Now consider the conjugation action of $P \in \operatorname{Syl}_2(G)$ on $\operatorname{Syl}_2(G)$. Since $N_P(Q) = P \cap Q$ (see, e.g., this answer), and the size of the $P$-orbit of any $Q \in \operatorname{Syl}_2(G)$ is $|P \cdot Q| = |P : N_P(Q)| = |P : P \cap Q|$. So, the action has $1$ fixed point ($P$ itself), and it partitions the set $\operatorname{Syl}_2(G) \setminus \{P\}$ of the remaining $14$ Sylow $2$-subgroups into orbits whose sizes are factors of $|P| = 2^5$ larger than $1$.

Since $2^2 \not\mid 14$, there is at least $1$ orbit of length $2$, hence some $R \in \operatorname{Syl}_2(G)$ such that $|P : P \cap R| = 2$; denote $T := P \cap R$. In particular, $T$ has index $2$ in both $P$ and $R$ and hence is normal in both. Since $N_G(T)$ contains both $P$ and $R$, $|N_G(T)| \geq 2^5 \cdot 3 = 96$, and hence $[G : N_G(T)] \leq \frac{480}{96} = 5$. Now, either $N_G(T) = G$, so that $T \trianglelefteq G$, which contradicts the simplicity of $G$, or $N_G(T)$ is a proper subgroup of index $\leq 5$, which contradicts our observation that every proper subgroup of $G$ must have index $\geq 8$.

$\mathbf{{}^*}$Remark In fact, if $G$ embeds in $S_k$, it must embed in $A_k$; otherwise $G \cap A_k$ would be an index-$2$---hence normal---subgroup of $G$, a contradiction. Moreover, Conway et al.'s ATLAS of Simple Groups shows that $A_8$ has no subgroups of order $480$, so in fact any proper subgroup of $G$ would have to have index $\geq 10$.

Travis Willse
  • 108,056
  • 1
    I don’t see why $\vert N_G(T) \vert \geq 64$. Could you please explain that conclusion? Is it because $2^5 \mid \vert N_G(T) \vert$ and also some odd prime $p \mid \vert N_G(T) \vert$ (which implies $\vert N_G(T) \vert \geq 96$)? – Robert Shore Jun 23 '23 at 00:11
  • 1
    It comes from looking at the largest possible overlaps of the Sylow-$2$ subgroups. Since $P, P'$ have a maximum intersection of size $2^4$, $P \cup P' \subset N_G(T)$ has at least $|P| + |P'| - |P \cap P| \geq 32 + 32 - 16 = 48$ elements. Now consider the third Sylow-$2$ subgroup of $N_G(T)$. (In fact, for this argument it's enough to see that $N_G(T)$ has $> \frac{|G|}{8} = 60$ elements.) But your observation is cleaner---I'll fold it into my answer. Thanks! – Travis Willse Jun 23 '23 at 00:44
  • 1
    At the start of your argument, $16$ distinct $5$-Sylow subgroups must account for $16 \cdot (5-1)=64$ elements of order $5$, not just $48$ such elements. The elements of the $2$-Sylow subgroups don't necessarily have order $2$, but they do have order that's a power of $2$, and that's good enough for your argument. – Robert Shore Jun 23 '23 at 02:19
  • You're right, thank you; I've fixed the issue. Of course, that adjustment does not affect the conclusion! – Travis Willse Jun 23 '23 at 02:22
  • 1
    Why is it necessarily the case that $P$ normalizes $T$? It's not obvious to me that $T \trianglelefteq P$, and this point (unlike the earlier nits) is essential to your argument. – Robert Shore Jun 23 '23 at 02:25
  • @RobertShore Hm, that conclusion needn't follow in general. It's sufficient to show that $P, P'$ are both of index $2$ (i.e., $|P \cap P'| = 2^4$), and that turns out to be true in this case as a consequence of the fact that $n_p \not\cong 1 \pmod{p^2}$ for $p = 2$. (That implication in turn follows from analyzing the conjugation action of $P$ on $\operatorname{Syl}_p(G)$.) But I was hoping to avoid using that lemma. I'll see if I can find another way to fix the gap, perhaps with a counting argument. Thanks for your close reading! – Travis Willse Jun 23 '23 at 03:03
  • (Any suggestions you have are also welcome.) – Travis Willse Jun 23 '23 at 03:08
  • (The noncongruence in my comment at 03:03Z should read $n_p \not\equiv 1 \pmod{p^2}$.) – Travis Willse Jun 23 '23 at 03:36
  • 5
    In the conjugation action of $P$ on the set of $15$ Sylow $2$-subgroups, there must be an orbit of length $2$, so we there is another Sylow $2$-subgroup $P'$ with $|P \cap P'|=8$. Now the normalizer in $G$ of $P \cap P'$ contains $P$ and $p'$, so it has order at least $3 \cdot 2^5$, which is is too big. – Derek Holt Jun 23 '23 at 08:10
  • @DerekHolt Your comment took me a bit to unpack, but I see now that it's a cleaner way of appealing to the ideas behind the lemma I mentioned above, and it helped clarify my thinking about the issue. Thanks for the especially helpful comment! – Travis Willse Jun 24 '23 at 02:31