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Timeline for embedding of $O_4^-(q)$ in $U_4(q)$

Current License: CC BY-SA 3.0

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Jan 6, 2016 at 23:37 vote accept Dima Pasechnik
Jan 6, 2016 at 16:10 answer added Nick Gill timeline score: 2
Jan 3, 2016 at 22:09 comment added Derek Holt $U_4(q)$ also had imprimitive sugroups isomorphic to $L_2(q^2)$, but I think they correspond to reducible subgroups of $\Omega^-_6(q)$.
Jan 3, 2016 at 22:00 comment added Dima Pasechnik @NickGill : thanks a lot, that's exactly what I hoped for! Care to convert your comment into a proper answer?
Jan 3, 2016 at 21:09 comment added Nick Gill (And, as Derek says, the definitive source on this is Kleidman and Liebeck, Section 4.3.)
Jan 3, 2016 at 21:07 comment added Nick Gill Given a $3$-dimensional quadratic form $Q$ over $\mathbb{F}_{q^2}$, any $\mathbb{F}_q$-linear function $L: \mathbb{F}_{q^2}\to \mathbb{F}_q$ will yield a $6$-dimensional quadratic form $L\circ Q$ over $\mathbb{F}_q$ (i.e. you don't need to use the trace form). What is more, you can get all possible $L$ from the trace by adding in some constant, i.e. $L(x)=L(\mu\cdot x)$ for some $\mu \in \mathbb{F}_{q^2}$. If you choose your $\mu$ carefully, the resulting linear form $L$ will give a minus type form $LQ$ for $q\equiv 3\pmod 4$.
Jan 3, 2016 at 17:32 history edited Dima Pasechnik CC BY-SA 3.0
tex fix
Jan 3, 2016 at 15:15 comment added Dima Pasechnik There seems to be more than one class of these, and figuring out what $H$ corresponds to looks tricky. And the value of $q\pmod 4$ seems to matter.
Jan 3, 2016 at 12:49 comment added Derek Holt Well $\Omega_n(q^2)$ ($n$ odd), or $\Omega^-_n(q^2)$ ($n$ even) is certainly a semilinear subgroup of $\Omega_{2n}^-(q)$ for all odd $q$ ($n$ odd) and all $q$ ($n$ even). I am sure you will find this all explained in Kleidman and Liebeck's book on the maximal subgroups of the finite classical groups.
Jan 3, 2016 at 11:53 history asked Dima Pasechnik CC BY-SA 3.0