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If there is a finite group with a normal subgroup and a representation of this subgroup over a finite field. How can one expand this representation to a representation of the whole group? Are there any conditions when this is possible?

Thanks a lot

edit: I am interesting above all in the case $G=D_n$ dihedral, $K$ finite field (e.j. $F_p$) and the subgroup $H$ is normal ( $[G:H]=2$)

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In this generality there is not much to be said. The normal subgroup could be just the identity. Also being a normal subgroup is not very relevant here, the problem can be posed for arbitrary subgroups. – Marc van Leeuwen Aug 28 '12 at 10:27
Do you know literature for this subject? Especially I try to do this for the dihedral goup $D_n$. – Eemeli Aug 28 '12 at 10:28
up vote 2 down vote accepted

Trying to addres the case you are particularly interested in, where you want to extend a representation of $C_n=\langle g\mid g^n=1\rangle$ to the dihedral group $D_n=\langle g,r\mid g^n=1=r^2, rgr^{-1}=g^{-1}\rangle$, and the field of scalars is the finite prime field $K=\mathbb{F}_p\simeq\mathbb{Z}/p\mathbb{Z}.$

A simple answer can be given, when $(n,p)=1$. From my answer to another question we extract the bit that an irreducible representation $M_{[a]}$ of $KC_n$ is fully determined by the cyclotomic coset $$[a]=\{ap^i\in\mathbb{Z}/n\mathbb{Z}\mid i\ \text{a non-negative integer}\}.$$ In the module $M_{[a]}$ a given generator $g$ of $C_n$ acts semisimply with eigenvalues (in a finite extension field of $K$) that are powers $\zeta^j, j\in[a]$, where $\zeta$ is a chosen primitive root of unity of order $n$.

In the dihedral group $D_n$ the generator $g$ is conjugate to its inverse $g^{-1}$, so a necessary condition for us to be able to extend $M_{[a]}$ to a representation of $D_n$ is that the element $g^{-1}$ acts with the same set of eigenvalues as $g$. Clearly this happens, if and only if $-a\in [a]$ (here $-a$ is also calculated modulo $n$). This condition is then also sufficient, because we can let the other generator $r$ act by intechanging the eigenvectors of $g$ belonging to eigenvalues $\zeta^j$ and $\zeta^{-j}$ for all $j\in[a]$.

Edit: More can be said. A useful looking observation is that when $-1\equiv p^i\pmod n$ for some natural number $i$, then $-a\in [a]$ for all the cyclotomic cosets $[a]$. When that happens, all the irreducible $KC_n$-modules can be lifted to representations of $KD_n$.

That settles the question in the case of irreducible representation. The above discussion implies that the reducible representation $M_{[a]}\oplus M_{[-a]}$ can always be extended to a representation of $D_n$. The general case requires that the irreducible summands of the representation can be paired up in this way.

This question also shows up in the theory of cyclic codes. It is possible to classify those cyclic linear codes that are also stable under the reversal of the symbols of codewords using this bit of theory.

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I think I got it now. Thank you all for your time! – Eemeli Aug 28 '12 at 11:19
Your edit seems interesting. But how to you get to this observation? – Eemeli Aug 28 '12 at 11:25
If $-1\equiv p^i\pmod{n}$, then $-a\equiv ap^i\pmod{n}$ for all $a$. Therefore the condition $-a\in [a]$ will be satisfied by all the cyclotomic cosets. – Jyrki Lahtonen Aug 28 '12 at 11:44

There is a general technique that allows to construct from a representation $\rho$ of any subgroup $H<G$ an induced representation $\text{Ind}_H^G(\rho)$ of the group $G$. The representation space is different (bigger), though.

But, when $[G:H]<\infty$, if $\text{Ind}_H^G(\rho)$ decomposes as a direct sum of $[G:H]$ copies of a representation $\rho^\prime$, then $\rho^\prime$ extends $\rho$.

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Thank you Andrea Mori, but when I induce a representation its dimension changes. I am looking for the cases when the dimension stays the same. This would be expansion instead of induction – Eemeli Aug 28 '12 at 10:49
@Eemeli : I just edited my answer adding a remark that might be a hint – Andrea Mori Aug 28 '12 at 10:50
Ok, in here I would have $[G:H]=2$. How could one decide, if $Ind_H^G(\rho)$ decomposes. Is it neccessary to "calculate" the induced rep.? – Eemeli Aug 28 '12 at 10:56
Given $\rho$, shouldn't be too hard to compute the character of the induced representation when $H$ has index $2$. What does Frobenius suggest? – Andrea Mori Aug 28 '12 at 11:01
Godd, I'll try it. Thank you – Eemeli Aug 28 '12 at 11:05

It is certainly not always possible to extend a representation from a subgroup (normal or not) to the whole group. For instance for the dihedral group $D_n$ one always has a $1$-dimensional complex representation of the cyclic subgroup $C_n$ of rotations, sending a generator to multiplication by an $n$-th root of unity. For $n>2$, this does not extend in any way to a $1$-dimensional complex representation of $D_n$, whose only $1$-dimensional representations are trivial on rotations.

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Thank you, but I need representations over finite fields, not complex. Are in this case (e.j. $G=D_n$, $K=F_p$) conditions when one can expand instead of induce the representation of the subgroup? – Eemeli Aug 28 '12 at 10:39
Oh, I missed that $K$ was finite. In any case, provided there are $n$-th roots of unity in $K^\times$, the situtation should be basically the same; the obstruction for extending to $D_n$ comes from the commutativity of $K^\times$. – Marc van Leeuwen Aug 28 '12 at 11:33

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