Let $G$ be a finite group, $n(G)$ the minimal number of generators and $m(G)$ the minimal number of irreducible complex representations generating (with $\otimes$ and $\oplus$) the left regular representation.

Notation: the word "generating" does not mean "generating exactly", but as a direct factor.

Question: Is it true that $n(G) \ge m(G)$ ?

Remark: a group $G$ is linearly primitive iff $m(G) = 1$, so it's obviouly true in this case.

  • $\begingroup$ What about $S_n$? It is generated by two elements; but I think, the minimum number of generators of irreducible representations is more than $2$. $\endgroup$
    – p Groups
    Jan 25, 2016 at 15:35
  • $\begingroup$ @pGroups: $S_n$ is linearly primitive because it is a primitive permutation group (see here), so $m(S_n) = 1$. $\endgroup$ Jan 25, 2016 at 15:47
  • $\begingroup$ Is linearly primitive equivalent to the group having a faithful irreducible representation? Clearly if it has such a rep, it is linearly primitive (as any irreducible is a summand of a suitable power of it), but I am unsure of the other direction. $\endgroup$ Jan 29, 2016 at 12:08
  • $\begingroup$ @TobiasKildetoft: my definition of linearly primitive is precisely the existence of a faithful irreducible complex representation. You are right, $G$ linearly primitive implies $m(G) = 1$. and the converse is true because if $m(G) = 1$ then the left regular representation $H = \mathbb{C}G$ is a direct factor of a combination (for $\otimes$ and $\oplus$) of one irreducible complex representation $V$. Now $H$ is faithful, so is $V$. $\endgroup$ Jan 29, 2016 at 12:18

1 Answer 1


Let $P$ be a $p$-group. Then the smallest number of irreducible characters of $P$ whose kernels intersect trivially is at least $n(Z(P))$, and thus $m(P) \ge n(Z(P))$. So to find an example where $n(P) < m(P)$, it suffices to find $P$ such that $n(P) < n(Z(P))$. An example in the Magma or GAP data base of small groups is SmallGroup($3^6$,9). Here, $n(P) = 2$ but $n(Z(P)) = 3$.

  • $\begingroup$ Do you know if there are precise results on the smallest number of irred. char. of a p-group P whose kernels intersect trivially ? In particular, your lower bound n(Z(P)) is exact if P has cyclic center. $\endgroup$ Feb 8, 2016 at 1:12
  • $\begingroup$ @ToddLeason: n(Z(P)) is exactly the minimum number of irreducible characters of a p-group P whose kernels intersect trivially. Choose that many linear characters of Z(P) whose kernels intersect trivially, and for each of them, choose one irreducible character of P lying over it. The kernels of these irreducible characters of P will have trivial intersection. $\endgroup$ Feb 8, 2016 at 4:19
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    $\begingroup$ @SebastienPalcoux Let E be the group of elements of order dividing p in Z(P). Then |E| = p^n, where n = n(Z(P)). Now if chi in Irr(G) has kernel K, then E/(E meet K) is cyclic, so has order at most p. Thus it requires at least n irreducible characters so that the intersection of their kernels meets E trivially. $\endgroup$ Feb 10, 2016 at 15:23
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    $\begingroup$ @SebastienPalcoux If chi is an irreducible character of an arbitrary finite group, then Z(G/K) is cyclic. In my previous answer, E/(E meet K) is isomorphic to EK/K, and this is contained in Z(G/K) since E is central in G. $\endgroup$ Feb 11, 2016 at 4:17
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    $\begingroup$ @yakov I think that what the questioner meant by speaking of a set of irreducible representations "generating" the regular representation is the following: Given a set S of irreducible characters of G, consider the set of all irreducible constituents of products of members of S, with repeated factors allowed. Then S "generates" if every irreducible character of G occurs as one of these constituents. The important point is that if S generates, then the intersection of the kernels of the members of S is trivial. $\endgroup$ Jun 28, 2016 at 19:15

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