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I'm working on question 22 of Dummit and Foote 18.1. I know this question has been answered in other posts, but I'm confused about the method this text recommends using:

Let $p$ be a prime, let $P$ be a $p$-group and let $F$ be a field of characteristic $p$. Prove that the only irreducible representation of $P$ over $F$ is the trivial representation. [Do this for a group of order $p$ first using the fact that $F$ contains all the $p$-th roots of 1 (namely 1 itself). If $P$ is not of order $p$. Let $z$ be an element of order $p$ in the center of $P$, prove that $z$ is in the kernel of the irreducible representation and apply induction to $P/<z>$.

So far, I think I've done the first part. I looked at a representation $\phi : P \to GL_n(F)$. I argued that the generator of $P$ is mapped to a matrix that satisfies $A^p = I$. So we know $A$ satisfies $x^p - 1 = (x-1)^p$. So 1 is the only eigenvalue and 1 is in F, so we put $A$ in Jordan-Canonical form (so has 1 on diagonal and on some parts of the super diagonal) and see that $[1,0,...0]^T$ is an invariant subspace. So if $n>1$, then we know $\phi$ is reducible and if $n=1$, then $A=1$ so we have the trivial representation. Is this logic okay? It honestly doesn't seem wholly correct to me.

For the second part, I'm stuck trying to show that $z$ is in the kernel of the irreducible representation. I'm assuming this is somehow tied to the fact that the $FG$-module affording the representation has only 0 and the whole module as submodules. But I'm getting stuck making the connections.

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Being a $p$-group, $Z(P)$ is not trivial, thus $Z(P)$ contains an element $g$ of order $p$. Just as your argument, there is a $0\neq x\in F^n$ such that $g\cdot x=x$. Then for any $g'\in P$, $g'\cdot x=g'g\cdot x=gg'\cdot x$. That is to say, all vectors in $\text{span}\{G\cdot x\}$ are eigenvectors of $g'$ corresponding to eigenvalue $1$. At the same time, $\phi$ being irreducible, the (nonzero invariant) subspace $\text{span}\{G\cdot x\}$ must equal to $F^n$. Thus $g$ acts trivially on $F^n$. Thus $\phi$ induces a irreducible representation of $P/(g)$ over $F^n$. Applying this process repeatedly, you will find the whole group acts trivially over $F^n$.

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