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The following problem comes from an algebra exercise and since two days or so, I am not able to find a satisfying solution:

Let $p$ be a prime with $p \geq 5$. Let $F_p$ denote the field with $p$ elements and let $G$ be the group $\operatorname{GL}(2,F_p)$. What are the composition factors of $G$?.

I already have the following idea: Let $S$ be the group $\operatorname{SL}(2,F_p)$ and consider the normal series

$\phantom{asasassasdssdsdssddssssssssssss}$ $G \geq S \geq Z \geq 1$

where $Z$ is the center of $S$. This cannot be a composition series since $G/S$ is a cyclic group of order $p-1$. However, a refinement of the above series which is a composition series must necessarily have the form

$\phantom{asasassasdssdsdssssssdds}$ $G \geq X_1 \geq ... \geq X_n \geq S \geq Z \geq 1$

where the $X_i$ are suitable subgroups of $G$. This follows from the facts that $Z$ is maximal normal in $S$ and that $Z$ is simple since $Z$ has order $2$. The question is of course how the groups $X_i$ should like or at least what orders they have. Up to now, I can only solve this problem for some special cases: if $(p-1)$ can be written as a product of two primes, say $q$ and $r$, it follows that the composition series has the form

$\phantom{asasassassssssssdssdsdssdds}$ $G \geq X \geq S \geq Z \geq 1$

where $|X| = |G|/q$ or $|X| = |G|/r$. Thus, the composition factors are

$\phantom{asasassassssssdssdsd}$ $\mathbb{Z}/q\mathbb{Z}$, $\mathbb{Z}/r\mathbb{Z}$, $\operatorname{PGL}(2,F_p)$ and $\mathbb{Z}/2\mathbb{Z}$

We even know that $q$ or $r$ must be $2$, so we can ignore one of first two cyclic groups and conlcude that $\mathbb{Z}/2\mathbb{Z}$ must have at least multiplicity $2$.

This is the state of my knowledge so far. However, I do not have any idea how to describe the composition factors in the general case.

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Since the question only asks for composition factors, you don't really need to identify what subgroups $X_i$ you'd like to use. Instead you just need to describe composition factors for (any, pick your favorite) cyclic group of order p-1.

Then, there is nothing special about p-1 being the product of two primes -- the composition factors are simply the primes dividing $p-1$. (You can see this for example from the classification of subgroups of cyclic groups, or -- if you prefer first principles -- pick a prime $q$ dividing $p-1$ and consider the (normal, since abelian) subgroup generated by $a^q$, then use induction on the group order.)

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  • $\begingroup$ I see. I didn't know the trick that in such a situation, it is sufficient to compute the composition factors of $G/S$ in order to complete the list of composition factors of $G$ (although it is intuitively clear that this should work). Thank you very much! $\endgroup$ – russoo Nov 26 '14 at 21:24

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