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Assume $G$ is a finite group and $H$ a subgroup. Is it true that the number of irreducible representations of $G$ is always larger than (or equal to) the number of irreducible representations of $H$?

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if $H$ is a normal subgroup, the answer is obviously Yes. –  Denis Serre Apr 4 '11 at 10:04
    
It can be equal: take $G=\frak S_3$ and $H=\frak A_3$. –  Denis Serre Apr 4 '11 at 10:06
    
The monster group has remarkably few characters given its size. –  Bruce Westbury Apr 4 '11 at 10:22
    
The simplest counterexample has order $10$. –  Tom Goodwillie Apr 4 '11 at 10:35
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Even if H is a normal subgroup, H can have more irreps than G. Take G to be dihedral of order 2n, and H to be cyclic of index 2. When n=4, G has 5 and H has 4. When n=3, G has 3 and H has 3. When n=5, G has 4 and H has 5. All three possibilities occur with normal subgroups. –  Jack Schmidt Apr 4 '11 at 16:37
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5 Answers 5

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Here is a simple example. As mentionned by Gjergji, this is a question about the number of conjugacy classes. Take $G=\frak A_4$, which has $3$ classes (the identity, the double transpositions and the $4$-cycles). Now take $H$ the subgroup spanned by a $4$-cycle. Because $|H|=4$ and $H$ is abelian, it has $4$ classes.

Edit. This is incorrect: $H$ is not a subgroup of $\frak A_4$, because a $4$-cycle is an odd permutation. I apologize. Note that I an accepted answer can not be deleted.

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The alternating group $A_4$ has four conjugacy classes. –  John Wiltshire-Gordon Apr 4 '11 at 14:19
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Your example is mostly fine, but the element description is off. A4 has three conjugacy classes (the identity, the double transpositions, and the 3-cycles) and 12 total elements. The subgroup H consisting of the identity and the double transpositions is a Sylow 2-subgroup of order 4. H has 4 irreducible representations and G has 3 irreducible representations. –  Jack Schmidt Apr 4 '11 at 16:26
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The three-cycles $(1 \, 2\, 3)$ and $(1 \, 2\, 4)$ are not conjugate in $A_4$. –  John Wiltshire-Gordon Apr 4 '11 at 16:43
    
@Jack. Thanks. I took it the wrong way. –  Denis Serre Apr 4 '11 at 16:43
    
@John: That is a problem! Yes, looks like A4 and H=K4 tie. In fact AΓL(1,q) and its normal elementary abelian subgroup of order q always tie, but for a very funny reason. There are q−2 conjugacy classes outside of the H, and only 2 inside the H. I think the dihedral groups (subgroups of AΓL(1,q) when q=n is prime) are probably the best to look at. They are varied at the beginning, but then clearly tend to H dominating. –  Jack Schmidt Apr 4 '11 at 16:51
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The number of irreducible representations equals the number of conjugacy classes, and the number of conjugacy classes in a subgroup may be more than in the whole group.

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Just thought of pointing out that the author of the above links is the following user of MathOverflow, mathoverflow.net/users/3040/vipul-naik –  Anirbit Apr 4 '11 at 17:04
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No, an easy example is when $G= {\bf Z}/2{\bf Z} \ltimes ({\bf Z}/p{\bf Z})^2$ where the action is by switching the factors and $H = ({\bf Z}/p{\bf Z})^2$.

$G$ has $2p$ 1dimrepresentations and $(p^2-p)/2$ 2dim representations which for big $p$ is smaller than the number of representations for $H$ which is $p^2$.

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Dihedral groups also give examples. More generally, for a subgroup $H$ of a finite group $G$, the sum of the $p$-th power of dimensions of representations of $H$ is at most that for $G$, provided $p\geq 1$ (here $p$ is not a prime, but one should think of $L^p$-norms, including the limit as $p$ goes to infinity, which corresponds to the maximal dimension.)

This monotony property fails for any $p$ in the interval $[0,1[$ (again, dihedral groups give examples).

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That's a nice result. Reference? –  John Wiltshire-Gordon Apr 4 '11 at 16:49
    
Sum( χ(1)^p ; χ in Irr(G) ) = Sum( (θ(1)[χ↓H,θ])^p, θ in Irr(H) ) ≥ Sum( θ(1)^p, θ in Irr(H) ), where the inequality uses the fact that every θ appears at least once in a χ, which follows by Frobenius reciprocity. When p is less than 1, then raising [χ↓H,θ] to the p'th power actually makes the sum smaller, so the inequality gets confused. –  Jack Schmidt Apr 4 '11 at 16:57
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In many cases, the result is rather in the opposite direction. Let $k(G)$ denote the number of conjugacy classes of a finite group $G$. Now take $G$ to be be a semdirect product of the form $VM$, where $V$ is an elementary Abelian $p$-group and $M$ is a group of linear transformations of order prime to $p$ of $V$. Then $k(G) \leq |V|$, and it is rare for equality to be achieved ( see Gluck, David; Magaard, Kay; Riese, Udo; Schmid, Peter The solution of the $k(GV)$-problem. J. Algebra 279 (2004), no. 2, 694–719). The precise conditions under which equality is achieved have not been characterized to date as far as I know), so taking $H = V$ usually leads to $k(G) < k(H)$ (and always to $k(G) \leq k(H)$) for this choice of $H$. For an explicit example, if $V$ is elementary Abelian of order $p^n$ for some odd prime $p$, and $M$ is cyclic of order $(p^n -1)/2$, acting as the square of a Singer cycle on $V$, then $G = VM$ has $2 + (p^n - 1)/2$ conjugacy classes, so this is less that $p^n$ unless $p^n = 3.$ While the result of Gluck,Magaard,Riese and Schmid and some of its prerequisites rely on the classification of finite simple groups, many cases can be proved without the classification, for example, the case where $G$ is solvable.

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