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Let K be a finite field and let R,P be groups (with R a subgroup of P). I know that the irreducible KP-modules have dimensions 1,4 and 16 over K. I have a KP-module M, and I know that M has dimension at most 5 over K. I also know that M does not have a quotient of dimension 1 over K. Moreover, if I consider M as a module over KR, it has a 2-dimensional irreducible module. Is it necessarily the case that M is a 4-dimensional irreducible module over KP? (my initial reasoning can be found in a comment below)

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The answer is no if and only if $\operatorname{Ext}^1(4,1)\neq 0$ in the obvious notation. If you said what P and R are maybe someone could work it out, but a priori it is not even obvious groups with these properties exist. – M T Feb 14 '12 at 14:02
Sorry, P and R are parabolic subgroups of the McLaughlin Group in a minimal parabolic system. – dward1996 Feb 14 '12 at 15:10
I think your answer has also proved that irrespective of the answer to the question in my specific case, there is a flaw in my reasoning. Thanks for the help. – dward1996 Feb 14 '12 at 15:24
Having looked at this again, I'm not sure that I follow the comment by mt. My initial reasoning was as follows. Suppose M were a 5-dimensional module. If M has a 4-dimensional submodule, then it must have a 1-dimensional quotient over K, which we know is not the case. Thus it must be the case that every irreducible submodule of M is 1-dimensional over K. However, the restriction of M to KR has a 2-dimensional irreducible submodule. A similar argument shows that M must have a 4-dimensional irreducible submodule, and thus as M has dimension at most 4, it is an irreducible 4-dimensional module – dward1996 Feb 20 '12 at 11:07
up vote 1 down vote accepted

I see no reason in general if $\operatorname{Ext}_P^1(4,1) \neq 0$ that, on restriction to R the four dimensional simple can't drop down and have a 2-dimensional submodule. I.e. On restriction to R it looks like:

\begin{equation}\begin{array}{c} 2\\ 2 \end{array} \oplus 1 \end{equation}

If you knew on restriction to KR the only irreducible submodule was two dimensional then this would be ruled out.

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Thanks for that. I'm not sure why I thought that my justification was correct! (Sorry!) – dward1996 Feb 21 '12 at 12:26

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