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Assume I have simple group G and its cyclic Sylow subgroup H. Are there any nice properties of the induced representation ?

Any remarks "around the situation" (relaxing "simple" "cyclic" "Sylow") are also welcome.

Example: Consider GL_3(F_2) which is remarkbly isomorphic to PSL_2(F_7).

It has cyclic Sylow subgroups of order 7 and 3. And irreps with dimensions 1,3,3,6,7,8.

$Ind_{Sylow 7} = V_1 \oplus V_7\oplus 2V_8$

$Ind_{Sylow 3} = V_1 \oplus V_{3a}\oplus V_{3b} \oplus 2V_6\oplus 3 V_7 \oplus 2V_8$

Here induction is from trivial character.

This is result of numerical calcultion with character table (hope I did not make mistake).

Is there some advice which can help me to predict the results in advance at least partitialy ?

I am making similar calculations with other groups, using character is a little painful, if something can be done more easily - would be very helpful.


Particular question: In particular I need to calculate inductions from non-trivial charcater in the cases above, any guess how to get the results without calculation ?

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Your induction from the Sylow 3-subgroup appears to be from a Sylow 2-subgroup; note that the dimension of the induced representation from the Sylow 3-subgroup should be 56, not 42. –  ARupinski Aug 18 '12 at 1:43
@Arupinski Yeh, you are right I copied from different line of my notes... sorry... I corrected: there should be 3 V_7, not 1V_7 as in previous –  Alexander Chervov Aug 18 '12 at 5:35
for myself: mathoverflow.net/questions/9154/… –  Alexander Chervov Aug 19 '12 at 12:16

2 Answers 2

up vote 8 down vote accepted

Here is a partial answer. (Your assumption that $G$ is simple is irrelevant for these remarks.) Suppose, as in your example, that the Sylow $p$-subgroup $P$ has order $p$ exactly. Then every irreducible character $\chi$ with degree divisible by $p$ appears as a constituent of $\lambda^G$ with multiplicity $\chi(1)/p$ for every irreducible character $\lambda$ of $P$.

There is a deep theory (due to Richard Brauer) which describes in some detail what the irreducible characters are of a group with a Sylow $p$-subgroup $P$ of prime order. To illustrate the power of the theory, I will indicate how to compute the multiplicity of an irreducible character $\chi$ in $(1_P)^G$ in the special case where $P = {\text C}_G(P)$. Write $e = |{\text N}_G(P):P|$, so $e$ divides $p-1$. Then except possibly for $(p-1)/e$ so-called $exceptional$ characters, each irreducible character of $G$ with degree not divisible by $p$ has degree congruent to $\pm1$ mod $p$. If $\chi(1) = kp + \varepsilon$, where $\varepsilon = \pm1$, then the multiplicity of $\chi$ in $(1_P)^G$ is exactly $k + \varepsilon$. The exceptional characters have equal degrees, and this common degree is congruent mod $p$ to $\pm e$. (Note that if $e = p-1$, there really is no exception.) If $\chi$ is exceptional with degree $kp + \varepsilon e$, where again $\varepsilon = \pm1$, then $\chi$ appears with mutiplicity $k$ in the induced character. This theory can also be used to find the multiplicity of $\chi$ in $\lambda^G$ for nonprincipal $\lambda \in \text{Irr}(P)$. For nonexceptional $\chi$, the answer is $k$, but the answer depends on the particular character $\lambda$ for the exceptional characters.

In fact, Brauer's theory has been generalized by Dade and others to the case where $P$ is cyclic, but is not of prime order. This is not the place to go into detail, however. One reference for much (but not all) of this theory is the book by G. Navarro from Cambridge University Press.

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Thank you very much, let me some time to think ! –  Alexander Chervov Aug 18 '12 at 5:37
Great answer ! Let me ask some questions to check my understanding... –  Alexander Chervov Aug 19 '12 at 12:17
Q1: You denote by $C_G(P)$ the commutant of subgroup $P$ in group $G$ is it correct ? –  Alexander Chervov Aug 19 '12 at 12:17
Q2: Take P=7-Sylow. V_3a, V_3b are exeptional ? And V_1, V_6,V_8 not ? Ind_{Sylow 7} = V_1 + V_7+ 2 V_8, corresponds to your answer like this: for V_1 we have 1=0*7+1 so k+$\epsilon$ = 1, so multiplicity is 1, and for V_6 we have +1-1=0 - so zero multiplicit, for V_8 we have +1+1 = 2 so multiplicity is 2. And about V_7 it is the first sentence 7/7=1. Is it correct ? –  Alexander Chervov Aug 19 '12 at 12:22
PS in Q2 I mean I consider my example of GL(3,2)=PSL(2,7) –  Alexander Chervov Aug 19 '12 at 12:22

This is a set of observations that should help you understand your inductions of non-trivial characters. If I read Marty's answer correctly, this should be a concrete way of understanding what he is saying using your particular example.

First note that each non-trivial character $\chi$ has a conjugate character $\overline{\chi}$ and that the inductions of $\chi$ and $\overline{\chi}$ must also be conjugate. Next, irreducible characters occur in sets which are related to one another by outer automorphisms of the dual group, these sets are defined by those irreducible characters with common kernel. Finally, note that the induction of the regular representation of the subgroup is the regular representation of the big group.

How does this help? Well in your example, lets look at what remains from the regular representation of $GL_3(\mathbb{F}_2)$ after we take away the induction of the trivial character of the Sylow 7-group. The remaining pieces are:

$3V_{3a} \oplus 3V_{3b} \oplus 6V_6 \oplus 6V_7 \oplus 6V_8$

Each induction of the 6 nontrivial characters must have dimension 24, and since the nontrivial characters are all more or less equivalent (up to outer automorphisms of the dual group to be technical), all their inductions should look more or less the same as well. This means that each induces to either $W = V_{3a}\oplus V_6 \oplus V_7\oplus V_8$ or its conjugate $\overline{W} = V_{3b}\oplus V_6 \oplus V_7\oplus V_8$.

With a bit of extra work, you would then find that if a character $\chi$ of the Sylow 7-subgroup induces to $W$ then so do the characters of $\chi^{\otimes 2}$ and $\chi^{\otimes 4}$ while the characters of $\chi^{\otimes 3}$, $\chi^{\otimes 5}$, and $\chi^{\otimes 6}$ will induce to $\overline{W}$, although the particular split on the exponents don't matter to your particular question (as currently stated).

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Thank you very much ! PS I corrected my post according to yours comment, thank you again ! –  Alexander Chervov Aug 18 '12 at 5:36
Yours answer is very helpful. If can accept two answers I would do it. –  Alexander Chervov Aug 19 '12 at 13:09

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