Michael Artin's Algebra, chapter 10 (both unstarred, and complex representations)

M.8 Prove that a finite simple group that is not of prime order has no nontrivial representation of dimension 2.

M.14 Let $\rho\colon G\to GL(V)$ be a two-dimensional representation of a finite group $G$, and assume that $1$ is an eigenvalue of $\rho_g$ for every $g$ in $G$. Prove that $\rho$ is a sum of two one-dimensional representations.

All these exercises are closely related to $GL_2(\mathbb C)$, and I think it's closely related to the property of $U_2$, the unitary group, therefore they go together.

We can simplify both questions in nearly the same way.

The first one:

It's not hard to show the correctness of abelian case, therefore we discard this case for now. Suppose there's a nontrivial 2D representation $\rho$ of a finite simple group $G$. Since $\rho$ is nontrivial and $G$ is simple, $\ker\rho$ is trivial, and $G$ embeds as a subgroup of $GL_2$. By Maschke's theorem, WLOG, we can suppose that $G\subset U_2$. Moreover, consider the mapping $\det\colon G\to\mathbb C$, we have $\ker\det$ is nontrivial, since $G$ isn't abelian, therefore by the normality of $G$, the image is trivial, and $G\subset SU_2$, the special unitary group.

The second one:

We can only consider the image of $\rho$. It's a finite group whose matrices have eigenvalue $1$. We'd only show that these matrices are simultaneously diagonalizable, therefore $\rho$ is a direct sum of two 1D representations. WLOG, suppose that the image is contained in $U_2$, by Maschke's theorem.

Both problems are simplified as a property of $U_2$ (the first one reduces a bit more). The first one says that there's no simple subgroup of composite number order, the second one says that if they all have eigenvalue $1$, then they're simultaneously diagonalizable.

How can we proceed? I need some insight of $U_2$ or $SU_2$. Thanks!

EDIT: I think my previous question is also related.

  • $\begingroup$ Should you being saying "irreducible representation" at a few points here? $\endgroup$ Aug 13, 2013 at 10:50
  • $\begingroup$ @AlexYoucis I don't see why the word irreducible should be applied anywhere. $\endgroup$ Aug 13, 2013 at 10:53
  • $\begingroup$ There is no abelian case in the first question; finite simple abelian groups are of prime order. $\endgroup$
    – anon
    Aug 13, 2013 at 11:05
  • $\begingroup$ Ok, so I am probably making the first one MUCH harder than it need be. Note that, for obvious reasons, your $G$ cannot have any non-trivial dim 1 reps. So, any non-trivial representation $\rho:G\to\text{GL}_2(\mathbb{C})$ must be irreducible and faithful. In particular, we may conclude that $2\mid |G|$ so that $G$ has an element $g$ of order $2$. Now, as you noted $\rho(g)$ has determinant $1$, but also has eigenvalues which satisfy $x^2=1$. In particular, it's a scalar matrix. $\endgroup$ Aug 13, 2013 at 11:33
  • $\begingroup$ But, this implies that $\rho(g)$ commutes with $\rho(x)$ for all $x\in G$, and since $\rho$ is faithful, this implies that $Z(G)\ne1$ which, by simplicity, implies that $G$ is abelian, which it isn't. Someone please tell me I'm speaking crazy things. $\endgroup$ Aug 13, 2013 at 11:34

4 Answers 4


Tobias Kildetoft has suggested that I add this as answer, but it still feels too hard:

M.8 Note that since $G$ is simple and non-abelian (else its of prime order) we know that any one-dimensional representation of $G$ is trivial. Now, suppose that $\rho:G\to\text{GL}_2(\mathbb{C})$ is non-trivial. Then, evidently $\rho$ is faithful. But, since it also can't be the sum of one-dimensional reps (since those are all trivial) it must also be irreducible. This implies, in particular, that $2\mid|G|$ so that $G$ has some element of order $2$, say $g$. Note then that $\rho(g)$ must have eigenvalues $\pm 1$ (since it's order $2$). But, as you've already noted, it must also have determinant $1$. This forces the eigenvalues to both be $-1$, and so, in particular, $\rho(g)$ is a scalar matrix, and so in the center of $\text{GL}_2(\mathbb{C})$. By faithfulness, this means that $g$ is a non-trivial element of $Z(G)$, which by simplicity forces $G=Z(G)$. This is a contradiction.

  • 1
    $\begingroup$ Is there any method to prove that $2\vert\#G$? Generally, if $\chi$ is an irreducible representation of $G$, $\chi(1)\vert\#G$ but it's left unproved on M.Artin's book. $\endgroup$
    – Yai0Phah
    Aug 14, 2013 at 9:51
  • $\begingroup$ The common technique is to show that $\displaystyle \frac{|G|}{\chi(1)}$ is an algebraic integer (by finding an integer valued matrix for which it is an eigenvalue), and then using the fact that the only algebraic integers which are rational are integers. It's in any standard book on rep theory. $\endgroup$ Aug 14, 2013 at 14:20
  • $\begingroup$ Well, next semester I'll try to read a proof from Serre's book. I hope it's helpful. According to the experience from doing other exercises on M.Artin, it seems that unstarred problems are usually easy, such as exercises on linear algebra where everything seems familiar, and I hope if there's an elementary argument on that. $\endgroup$
    – Yai0Phah
    Aug 14, 2013 at 14:51

M.14: We may assume that $\rho$ is faithful (just replace $G$ by $G/\ker(\rho)$ and neither the hypothesis nor the conclusion changes). In particular, no non-identity element is in the kernel of $\det$ since $\det(g)$ is precisely the non-one eigenvalue of $g$. Thus $\det:G/\ker(\rho) \to \mathbb{C}^\times$ is an embedding, and $G/\ker(\rho)$ is finite abelian, so every representation is a direct sum of one-dimensional representations.

  • $\begingroup$ A nice trick, thanks! $\endgroup$
    – Yai0Phah
    Aug 13, 2013 at 15:45
  • $\begingroup$ Though the proof is short, it's no easy for me to derive. $\endgroup$
    – Yai0Phah
    Aug 13, 2013 at 15:47
  • $\begingroup$ @Jack Schmidt Can you see an easier solution for M.8? $\endgroup$ Aug 13, 2013 at 18:27
  • $\begingroup$ @Alex I think your method is standard and important. Calculating $Z(\rho)$ especially elements of order 2 is useful. $\endgroup$ Aug 13, 2013 at 19:34
  • $\begingroup$ @Frank I like the trick because it is the opposite of the trick you used in M8. $\endgroup$ Aug 13, 2013 at 19:35

Another solution for M.14 (this is probably easier for you to come up with than Jack's answer):

Suppose in contradiction that $\rho$ is not a sum of two one-dimensional representations, it must be irreducible. Since one of the eigenvalues is 1, the matrix for $\rho(g)$ must look like $$\begin{pmatrix}1 & 0\\\ 0 & \lambda_g\end{pmatrix}$$hence $\chi(g) = 1 + \lambda_g$.

Now, take the inner product of $\chi$ with the unit representation. (As a side note, this is a standard trick in similar exercises where you know something about the sum of the characters.) Because a two dimensional representation can never be isomorphic to a one dimensional representation, this inner product must equal to zero (as they are both irreducible). Thus $$\frac{1}{|G|}\sum_{g \in G} \chi(g)= 0 = 1+\frac{1}{|G|}\sum_{g \in G} \lambda_g$$ By the triangle inequality, $$|\sum_{g \in G} \lambda_g| \le \sum_{g \in G} |\lambda_g| = |G| $$ with equality taken when all $\lambda_g$'s are equal and as the sum is -|G|, they would all have to be -1. But the norm of this representation is 0 $\neq$ 1, so our assumption fails and $\rho$ should not be irreducible, hence a sum of two one-dimensional representations.


As Frank notes for M8, $G \leq SU_{2} (\mathbb{C})$. There is a 2-to-1 group homomorphism: $$SU_{2} (\mathbb{C}) \to SO_{3} (\mathbb{R})$$ This is the universal cover of $SO_{3} (\mathbb{R}) \cong \mathbb{R}P^{3}$ (Artin covers this-I believe- in chapter 9). Since $G$ is simple, this means $G \leq SO_{3} (\mathbb{R})$ also, which means that $G$ is a finite subgroup of what Artin calls the ``rotation group" (at the end of chapter 6) meaning a group acting faithfully on the sphere via rigid motion. Artin classifies all such finite groups. There are only five possibilities for $G $ according to this classification:

  1. $\mathbb{Z}/n$

  2. The dihedral group $D_{n}$

  3. The tetrahedral group $A_{4}$

  4. The octahedral group $S_{4}$

  5. The icosahedral group $A_{5}$

The only group in this list which is not of prime order and is simple is $A_{5}$, which has 5 irreducible complex representations. One of these is the trivial one and the remaining 4 have dimension > 2. So, it has no non-trivial representations of dimension 2. Hence, M8. This may have been what Artin had in mind, although I like Alex's argument better!

  • $\begingroup$ Does this answer the question? $\endgroup$
    – Qwerty
    Jun 28, 2016 at 12:40

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