# Representation of a group of prime power order

Let $G$ be a group, $|G| = p^t$, $p$ prime, $t \geq 2$. It is easily seen that $\frac{G}{[G,G]}$ has order at least $p^2$. Now, I want to prove that every irreducible representation of a group of order $p^4$, has dimension $1$ or $p$.

Clearly, such a representation must have $\dim \rho < p^2$, since $\sum_{\rho \in \operatorname{Irr}(G)} \dim(\rho)^2 = |G|$, and there is at least the trivial representation of order $1$, so my claim follows. Now, with the additional fact that $\dim \rho | \ |G|$, I am done. However, I am asked not to use this.

I do know the existence of representations of order $1$ and $p$, however ruling out any other dimension is not clear to me.

I know basic group theory : say, up to Sylow's theorem related arguments, and representation theory up to the basics, which is Wigner's little groups method.

I would not mind seeing proofs which are outside this prescription, however I would prefer proofs which are more elementary.

• With questions like this you really need to provide more information on exactly what you are allowed to use. Can you use Clifford's theorem for example? – Derek Holt Nov 6 '17 at 9:04
• @DerekHolt I read up Clifford's theorem, and it's proof seems to use projective representations, although I could be wrong. But I have made an edit above : even if the proof is not easy, include it, that can only be helpful to me. Only do not use directly the simple fact that the dimension of an irreducible representation divides the order of the group. My hunch was, that with a group of order a small prime power , and some "Sylow's theorem" related arguments, maybe we can come up with an elementary proof rather than one using more machinery. Again, I could be wrong. – астон вілла олоф мэллбэрг Nov 6 '17 at 10:40

If $G$ is abelian then all irreducible representations have degree $1$, and the result follows. So assume that $G$ is nonabelian.

We show first that there is an abelian normal subgroup $N$ of $G$ of order $p^3$. To see this, let $K$ be a normal subgroup of order $p^2$, so $K$ is abelian. Then $G/C_G(K)$ is isomorphic to a subgroup of ${\rm Aut}(K)$. We have either $K\cong C_{p^2}$ and $|{\rm Aut}(K)| = p(p-1)$, or $K \cong C_p \times C_p$ and $|{\rm Aut}(K)| = p(p^2-1)$. So in either case $|{\rm Aut}(K)|$ is not divisible by $p^2$, and hence $|C_G(K)| \ge p^3$, and we can take $N$ to be a subgroup of $C_G(K)$ of order $p^3$.

Now Clifford's Theorem says that, if $\rho$ is an irreducible representation of $G$ and $N \unlhd G$, then the restriction $\rho_N$ of $\rho$ to $N$ is equivalent to the sum of representations $\rho_1 \oplus \cdots \oplus \rho_k$ for some $k$, where each $\rho_i$ is itself the sum of isomorphic irreducible representations $\rho_{i1} \oplus \cdots \oplus \rho_{it}$ for some $t$, and where $\rho_{ij}$ is equivalent to $\rho_{i'j'}$ iff $i=i'$. Furthermore, the representations $\rho_i$ are permuted transitively by the conjugation action of $G$.

We apply this to our abelian normal subgroup $N$ of order $p^3$, and assume that $\rho$ has degree greater than $1$. Since $N$ is abelian, the $\rho_{ij}$ all have dimension $1$. If $k=1$ then the irreducible constituents of $\rho_N$ are all equivalent, so the image of $\rho_N$ consists of scalar matrices. But then $N/K \le Z(G/K)$, where $K$ is the kernel of the representation. But since the quotient by the centre cannot be cyclic and nontrivial, this implies that $G/K$ is abelian, and hence $\rho$ has degree $1$, contrary to assumption. So $k>1$. Since the $\rho_i$ are permuted transitively by $G$, $k$ must divide $|G| = p^4$, and since $\rho$ has degree less than $p^2$, we must have $k=p$.

But since $|G/N|=p$ and $\rho$ is irreducible, $\rho_N$ cannot have more than $p$ irreducible constituents, and hence $t=1$ and $\rho$ has degree $p$.

So we have proved that all irreducible representations of $G$ have degree $1$ or $p$.

• I was relatively close :-) I got stuck with the possibility that $G'=[G,G]$ has order $p$, and couldn't rule out the possibility that $C_G(G')$ is isomorphic to the group of upper triangular 3x3 matrices over $\Bbb{F}_p$. What I missed was the possibility to start with some normal subgroup of order $p^2$ :-( – Jyrki Lahtonen Nov 6 '17 at 14:22
• Just a note that you can do this without Clifford theory. Indeed, once you have an abelian subgroup $A$ (not necessarily normal), all irreducible characters have degree $\le[G:A]$. This follows by looking at an irreducible representation of $G$, and restricting it to $A$. The image has a one-dimensional irreducible subrepresentation, and its "cosets" generate the original vector space. – Steve D Nov 6 '17 at 16:18
• @DerekHolt: Not sure I follow? Are you referring to the condition in the OP of not using that $\chi(1)$ divides $|G|$? – Steve D Nov 6 '17 at 16:30
• @SteveD By applying your argument to a subgroup $A$ of index $p$, we find that all irreducibles have degree at most $p$. I was wondering how you were going to show that the nonlinear irreducibles have degree at least $p$. – Derek Holt Nov 6 '17 at 16:48
• @DerekHolt: Yes, my intention was that we could use $\chi(1)$ divides $|G|$, but it seems the OP disallows this. – Steve D Nov 6 '17 at 18:21

Here is a proof not using Clifford theory. All arguments are taken from Chapter 2 of Isaacs's book on character theory, in particular the following facts (see results 2.27--31 and Exercise 2.9(b) there). $\DeclareMathOperator{\Irr}{Irr}$

Fact 1. Let $\chi$ be a character of a finite group $G$ and $H\leq G$ a subgroup. Then $$[\chi_H, \chi_H] \leq \lvert G:H \rvert [\chi,\chi]$$ with equality iff $\chi$ vanishes on $G\setminus H$.
($\chi_H$ is the restriction of $\chi$ to $H$ and $[.,.]$ is the usul inner product on class functions.)

Proof. Clear from $$\lvert H \rvert [\chi_H, \chi_H ] = \sum_{h\in H} \lvert \chi(h) \rvert^2 \leq \sum_{g\in G} \lvert \chi(g) \rvert^2 = \lvert G \rvert [\chi,\chi].$$

Fact 1a. Let $\chi \in \Irr(G)$ and $A\leq G$ an abelian subgroup. Then $\chi(1) \leq \lvert G : A \rvert$.

Proof. For $A$ abelian, we have $$[\chi_A, \chi_A] = \sum_{\lambda} [\chi_A, \lambda]^2 \geq \sum_{\lambda} [\chi_A, \lambda] = \chi(1).$$ Now use Fact 1.

Fact 2. Let $\chi\in \Irr(G)$ and $Z=Z(G)$, the center of $G$. Then $\chi_Z = \chi(1) \lambda$ for some linear character of $Z$, and $\chi(gz)=\chi(g)\lambda(z)$ for all $g\in G$, $z\in Z$.

Proof. The irreducible representation affording $\chi$ acts by scalars.

Fact 3. Let $\chi \in \Irr(G)$ be faithful and suppose that $g$, $h\in G$ are such that $1 \neq [g,h] (:=g^{-1}h^{-1}gh) \in Z(G)$. Then $\chi(g)=\chi(h)=0$.

Proof. $\chi(g)=\chi(g^h) = \chi(g[g,h]) = \chi(g)\lambda([g,h])$, with $\lambda$ as in Fact 2. Since $\chi$ is faithful, we have $\lambda([g,z])\neq 1$, and thus $\chi(g)=0$. Since $[g,h]^{-1} = [h,g]$, also $\chi(h)=0$.

Answer to question: Suppose that $G$ is a $p$-group with $\lvert G \rvert \leq p^4$ and $\chi\in \Irr G$ is faithful. (We can always first factor out the kernel of $\chi$.) Let $Z=Z(G)$ and $X/Z = Z(G/Z)$, so $X$ is the second term in the ascending central series. When $Z=G$, then $G$ is abelian, so $\chi(1)=1$. Otherwise, for every $x\in X\setminus Z$, there is $g\in G$ with $1 \neq [x,g]$. But also $[x,g]\in Z$, and so $\chi(x)=0$ (Fact 3). Thus $\chi$ vanishes on $X\setminus Z$. From Fact 1 applied to $Z\leq X$ it follows $$\chi(1)^2 = [\chi_Z,\chi_Z] = \lvert X : Z\rvert [\chi_X, \chi_X].$$
When $X=G$, this yields $\chi(1)^2 = \lvert G : Z \rvert$, and we are finished.
Otherwise, we see at least that $p$ divides $\chi(1)$. In this case, we must have $\lvert G \rvert = p^4$, $\lvert X \rvert = p^2$. Then there is an abelian $C$ with $X < C < G$. (As in Derek Holt's answer, or observe that for any $x\in X\setminus Z$, the centralizer $C=C_G(x)$ must be a proper subgroup of order $p^3$, because $[x,G]\subseteq Z$. Because $\langle x, Z\rangle \subseteq Z(C)$, this $C$ must be abelian.) Then $\chi(1) \leq p$ by Fact 1a.