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Timeline for On the groups of order $p(p^2+1)$

Current License: CC BY-SA 3.0

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May 8, 2014 at 17:19 vote accept BHZ
May 8, 2014 at 13:57 comment added Geoff Robinson Yes, marginally easier. I had forgotten that $P$ was normal, so Burnside's transfer theorem certainly wasn't necessary.
May 8, 2014 at 13:56 comment added Peter Mueller I guess a slightly easier argument, using the normality of $P$, is: $L/\text{C}_L(P)$ is a subgroup of $\text{Aut}(P)=C_{p-1}$. But $\text{gcd}(p-1,\lvert L\rvert)=2$, so $[L:\text{C}_L(P)]\le2$.
May 8, 2014 at 13:34 history answered Geoff Robinson CC BY-SA 3.0