I have posted this question on mathstack echange but did not get any answer. It mam trying my luck here.

The only simple finite groups admitting an irreducible character of degree 3 are $\mathfrak{A}_5$ and $PSL(2,7)$. That seems to be a result coming from Blichfeldt's work on $GL(3,\mathbb{C})$, which I cannot find. Is there a proof available somewhere ?

  • $\begingroup$ This is also done in Determination of the ordinary and modular ternary linear groups Howard H. Mitchell. Trans. Amer. Math. Soc. 12 (1911), 207-242, but that is a long and complicate paper. Why can't you find Blichfeldt's papers? Do you not have access to Trans Amer Math Soc? It's an interesting question, because all of the modern treatments that I have found seem to cite Blichfeldt or Mitchell. $\endgroup$ – Derek Holt Jul 7 '16 at 20:06
  • $\begingroup$ Thanks Derek. I found Mitchell's article. To answer your question, I only have access to articles freely available on the internet. $\endgroup$ – Nicolas Jul 7 '16 at 21:07
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    $\begingroup$ The articles of H.F. Blichfeldt in the AMS Transactions are indeed freely available on the internet: ams.org/epubsearch/servlet/… $\endgroup$ – Jim Humphreys Jul 7 '16 at 22:09
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    $\begingroup$ It might be worth remarking that the triple cover of $A_{6}$ ( which is not simple of course, but is quasi-simple) has a complex irreducible character of degree $3$. $\endgroup$ – Geoff Robinson Jul 8 '16 at 3:36
  • $\begingroup$ See \S 8.5 in W. Feit, The current situation in the theory of finite simple groups, Actes. Cobgr. Internat. Me\ath. Nice 1970, vol.1 Gauthhier-Villars, Paris, 1971, 55-93. $\endgroup$ – yakov Jul 8 '16 at 13:06

It depends on how much group theory you want to use. If $G$ is such a simple group and $\chi$ is a faithful complex irreducible character of degree $3$, then a Theorem of Feit and Thompson proves that $|G|$ is not divisible by any prime $p > 7$. It is easy to check (since $Z(G)$ contains no element of order $3$), that $G$ has Abelian Sylow $3$-subgroups. Hence for any $x \in G$ of order $3$, we have ${\rm gcd}([G:C_{G}(x)],\chi(1)) = 1$. By a result of Burnside, we have $\chi(x) = 0$ since $G$ is simple. It then easily follows that $|G|$ is not divisible by $9$. It also follows that a Sylow $3$-subgroup of $G$ is self-centralizing ( from this, it already follows from 1962(?) theorem of Feit and Thompson in Nagoya J. Math, that $G \cong {\rm PSL}(2,7)$ or $A_{5}$. For, more generally, we see that $\chi(g) = 0 $ whenever $g$ centralizes a Sylow $3$-subgroup $P$ of $G$, and then $|C_{G}(P)|$ divides $3$). A representation theoretic proof in this special case can be sketched as follows (using a few tricks not available to Blichfeldt):

Now $G$ has cyclic Sylow $7$-subgroups for otherwise, $G$ contains a element of order $7$ with eigenvalues $1,\omega, \omega^{-1}$, where $\omega = e^{\frac{2 \pi i}{7}}$, which contradicts a theorem of Blichfeldt ( as $\chi$ must be primitive ( otherwise $G$ would be solvable)). Consideration of reduction (mod $7$) shows that the Sylow $7$-subgroup of $G$ has order at most $7$. Furthermore, if $7$ divides $|G|$, it follows from Burnside's normal $p$-complement theorem that a Sylow $7$-normalizer must have order $21$.

Since $G$ has no element of order $35$, it follows from another theorem of Blichfeldt that $|G|$ can't be divisible by $35$ ( since otherwise, $\chi$ is neither $5$-rational nor $7$-rational, and would contain an element of order $35$. But consideration of reduction (mod $5$) shows that no element of order $5$ can commute with any element of order $7$).

If $G$ contains an element of order $5$ with only two eigenvalues then an argument of Blichfeldt shows that ${\rm SL}(2,5)$ as a subgroups, and contains an element of order $6$ with eigenvalues $1, \alpha, {\bar \alpha}$ for a primitive $6$-th root of unity, which ( by another of his results) contradicts the primitivity of $\chi$. It follows that $G$ has cyclic Sylow $5$-subgroups which have order at most $5$ on consideration of reduction (mod $5$).

It now follows that $|G|$ has the form $2^{a}.3.5$ or $2^{b}.3.7$ for integers $a,b$. Similarly to the argument for $7$, we may conclude that a Sylow $5$-normalizer has order $10$ if $5$ divides $|G|$. This gives $a \equiv 2$ (mod $4$) or $b \equiv 3$ (mod $6$) by Sylow's theorem. A Sylow $2$-subgroup $S$ of $G$ has an Abelian normal subgroup $A$ of index dividing $2$, and a 1965 Theorem of Brauer shows that $|A|$ divides $4$ so $|S| \leq 8$ and we do get $|G| = 60$ or $168$.

  • $\begingroup$ Thanks vm Geoff. It will take me some time to go through your proof but it is extremely precious ! May I just ask: when you mention reduction, do you refer to the use of modular representations or something else ? $\endgroup$ – Nicolas Jul 7 '16 at 21:06
  • $\begingroup$ Yes, reduction mod p refers to realising the representation over a local ring and looking at images mod the unique maximal of that ring in the standard manner. $\endgroup$ – Geoff Robinson Jul 7 '16 at 22:06
  • $\begingroup$ @ Geoff. Do you have a preferred reference for some examples of reduction? Also, how do you prove that a Sylow $-$normalizer must have order 21. All I have is $C_G(P<N)G(P)$ by Burnside and $[N_G(P):C_G(P)]\in \{2,3,6\}$ by the N/C lemma. $\endgroup$ – Nicolas Jul 8 '16 at 15:10
  • $\begingroup$ The Sylow $7$-subgroup is self-centralizing by a subtle argument which I think originates with Brauer. Take a prime $q \neq p$ and suppose $x \in C_{G}(P)$ is an element of order $q$. Reduce the representation (mod $q$), and note that $P$ is diagonalizable, and since a generator of $P$ has three different eigenvalues, anything which centralizes $p$ is daigonalizable too. Hence $x$ acts trivially (mod $q$), so $x \in O_{q}(G) = 1$. Alsao $N_{G}(P)$ can't be a Frobenius group of order $42$. I have still left out a few details. $\endgroup$ – Geoff Robinson Jul 8 '16 at 18:16
  • $\begingroup$ As for reduction $(mod $p$)$, I like Curtis and Reiner 1962 ( Representation Theory of Finite Groups and Associative Algebras) but all books on modular representation theory cover it. $\endgroup$ – Geoff Robinson Jul 8 '16 at 18:17

A comprehensive report on groups possessing a faithful irreducible character of small degree is contained in Walter Feit's report `The current situation in the theory of finite simple groups, Actes Congr. Internat. Math. Nice 1970, vol. 1 Gauthier-Villars, Paris, 1971, 55-93.'

  • $\begingroup$ How does this answer the question ? This is at best a comment. $\endgroup$ – Todd Leason Jul 12 '16 at 14:05
  • $\begingroup$ That report by Feit contains a number of inaccuracies. $\endgroup$ – Dima Pasechnik Jul 12 '16 at 14:56

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