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I have no issue with finding representations over $\mathbb{C}$ but I'm not quite sure how to find them over finite fields, particularly when that field might not be algebraically closed or the characteristic of the field divides the order of the group (both of which are possible in the above question). Thank you in advance for your help.

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    $\begingroup$ Dear john, Have you first considered the case $p > 3$. Then $\mathbb F_p$ has char. prime to $|A_4|$, and so if you work over $\overline{\mathbb F}_p$, the story is the same as for $\mathbb C$. Now the reps. of $A_4$ are pretty concrete, and you could think about whether or not you can define them over $\mathbb F_p$. (E.g. when can you define a non-trivial character of order $3$ over $\mathbb F_p$?) Regards, $\endgroup$ – Matt E May 28 '13 at 2:14
  • $\begingroup$ Hi Matt, thank you for your response. Sadly my knowledge of field theory is a bit limited. It seems to me that since over $\mathbb{C}$ the characters of $A_4$ require third roots of unity, we could find the usual representations over $\mathbb{F}_p$ if this field has third roots of unity and since the multiplication group of $\mathbb{F}_p$ is cyclic with order p-1, it seems this is the case if 3 divides p-1. If 3 doesn't divide p-1 though that's when things seem to get a bit weird. (continued in next comment) $\endgroup$ – john May 28 '13 at 5:02
  • $\begingroup$ I know for instance when finding representations of the cyclic group of order three over $\mathbb{F}_5$ we can get an irreducible two dimensional representation (as opposed to three one dimensional representations). My current knowledge of field theory doesn't extend to $\bar{\mathbb{F}}_p$ yet (though I will read up more on this) so perhaps this method could be the key to determining the representations when 3 is not a divisor of p-1. If you have the time, an example of using this method could be very helpful. Thanks again. $\endgroup$ – john May 28 '13 at 5:07
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Here is a quick answer. We get the deleted permutation module of dimension 3, which is irreducible over all fields of odd characteristic.

This is the only faithul irreducible representation. The others all have the subgroup of order 4 in their kernels, so they are effectively representations of the cyclic group of order 3.

We always have the trivial representation, and that is the only other one when when $p=3$.

When $3|p-1$, there are two faithful 1-dimensional representations. (Remember that ${\rm GL}_1(p)$ is just the cyclic group of order $p-1$.)

Otherwise, we have a single irreducible but not absolutely irreducible 2-dimensional representation, which decomposes into two 1-dimensionals over the field of order $p^2$. In this case, you can think of the cyclic group of order 3 as a subgroup of ${\rm GL}_1(p^2)$.

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  • $\begingroup$ Hi Derek (or others). A thought just occurred to me. In the case of a non-algebraicly closed field, how do we know that the representations lifted from the cyclic group of order 3 are the only representations other than the faithul irreducible representation? In the algebraicly closed field we know how many representations to expect so once we've found that number we're done. In this other case though, how do we know there aren't some other hiding out there somewhere? Thank you $\endgroup$ – john Jun 1 '13 at 13:03

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