TITLE: Representation of a group of prime power order
QUESTION [7 upvotes]: 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.

REPLY [2 votes]: 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.