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Let S be a finite set of integers, do I can check with Gap that this set be a set of character degrees of small group?

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No one has mentioned yet: the order of the group must be a multiple of the LCM of the elements of S. – John Wiltshire-Gordon Oct 13 '12 at 19:38
up vote 4 down vote accepted

Dima is correct that the problem is much easier with multisets. Furthermore, taking direct products with Abelian groups, it is clear that for any given set of irreducible character degrees which actually occurs, there are infinitely many groups for which it occurs, if we allow multiplicities. I had a recollection that Isaacs and Passman proved a kind of converse of this: if a finite group $G$ has all its irreducible character degrees of degree less than $d,$ then $G$ has an Abelian normal subgroup $A$ whose index is bounded in terms of $d.$ I could not locate a reference at the time, though I did not have Isaacs' book to hand. Since my original post, however, Yemon Choi kindly provided a reference in the comments below (see MR0167539). The wording of Isaacs and Passman's result is perhaps mildly ambiguos: they speak of an Abelian subgroup of index bounded in terms of $d$, omitting the word "normal", but of course if $G$ has an Abelian subgroup $A$ with $[G:A] \leq h,$ then the intersection of the $G$-conjugates of $A$ is normal, and has index at most $h!$. Hence, in general, if we know the irreducible character degrees of a finite group $H,$ there is an Abelian normal subgroup $A$ of $H$ such that $|H/A|$ is bounded in terms of the largest degree. So in small cases, it may be possible to use GAP to determine all possibilities for $H/A.$

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I am out of the office right now so cannot view it, but I think the result of Isaacs and Passman that you refer to is in their paper Groups with representations of bounded degree, Canad. J. Math. 16 (1964) 299--309. – Yemon Choi Oct 13 '12 at 23:58
Thanks, that figures- I put a date range in Mathscinet starting at 1965! – Geoff Robinson Oct 14 '12 at 1:24

Given an arbitrary set $S$ of powers of some prime $p$, subject only to the condition that $1 \in S$, there necessarily exists a $p$-group $P$ such that $S$ is exactly the set of irreducible character degrees of $P$. In fact, it is always possible to find $P$ having nilpotence class $2$. This theorem appears in my paper in PAMS 96 (1986).

There is no analogous result for arbitrary sets of integers. For example, no group has irreducible character degree set $\{1,2,3,5\}$.

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You most probably mean "the set of degrees of irreducible characters". Indeed, given a finite set of integers, it should not be hard to construct a finite group that has irreducible characters of that degrees (and many more other degrees).

Still, for "the set of degrees" the problem does not look easy, as degrees can repeat, and you don't have a good way to control the order of the group $G$. For "the multiset of degrees" $d_1$,...,$d_t$ things are much better, as then $G$ has order equal to $\sum_{j=1}^t d_j^2$, and so you can just look through GAP's database of small groups for such a group.

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