I have trouble with the following exercise:
$\textbf{Exercise}$ Let $H$ be a finite $p$-group, let $A \trianglelefteq H$ be a normal subgroup which is an elementary abelian $p$-group and let $x \in H$ such that $[A,x^{p^n}] \neq 1$ for some $n$. Prove that there exists an element $a \in A$ such that $[a,x^{p^n}]=1$ and $\langle a,x \rangle/_{\langle x^{p^n}\rangle}\cong C_p \wr C_{p^n}$.
Where given $X$ a finite group $C_p\wr X=C_p[X]\rtimes X$ with $X$ acting on $ C_p[X]$ by right multiplication.
We say a group $G$ involves another group $K$ iff there exist $M \trianglelefteq N \leq G$ such that $N/M \cong K$.
I managed to prove the existence of $a$ as follows: we can consider $A$ as a vector space $(\mathbb{F}_p)^r$, then for every $h \in H$ the conjugation map by $h$ gives us a linear automorphism of $A$ since it is a normal subgroup. We define $\lambda$ to be the automorphism associated to $x^{p^n}$. The condition $[A,x]\neq 1$ means $\lambda$ is not the identity and $[a,x^{p^n}]=1$ can be rephrased as $a$ being an eigenvector of $\lambda$ with eigenvalue $1$.
Thus I have to show $\lambda$ admits as eigenvalue $1$ and I proceed as follows. Since $x^{p^n} \in H$ a finite $p$-group it must have order a power of $p$, say $p^i$. This implies $\lambda^{p^i}=1$ i.e. $0=\lambda^{p^i}-1=(\lambda-1)^{p^i}$. Since the minimal polynomial must divide the last expression we deduce it is in the form $(x-1)^{p^j}$ for $j \leq i$. Therefore it admits as solution $1$ hence $1$ is an eigenvalue of $\lambda$.
At this point I have two problems:
1) since I proved the minimal polynomial is in the form $(x-1)^{p^j}$ its only possible solution is $1$, so the only eigenvalue should be $1$ implying that $\lambda$ is the identity, which is against the assumption. Where is my mistake?
2) I do not understand how to provide the isomorphism $\langle a,x \rangle/_{\langle x^{p^n}\rangle}\cong C_p \wr C_{p^n}$. My idea was to just find in $C_p \wr C_{p^n}$ two explicit generators corresponding to $a$ and $x$, but this is much more difficult that I expected. The problem is that one would expect that $x$ to correspond to $(0,1)\in C_p[C_{p^n}]\rtimes C^{p^n}$ or something similar but then when imposing commutativity with another $(\sum_{i \in C_{p^n}}a_i i, \sigma)$ you get the condition $a_i=a_j$ for all $i,j \in C_{p^n}$ (or something similar). And it is easy to see they cannot generate the wreath product. Should I adopt a different approach entirely? I do not have a lot of experience with wreath products, so I do not know if writing down these two generators is something feasible or you can show only theoretically that they exist.
The action of conjugation by $x$ on $A$ regarded as vector space over the field of order $p$ induces a linear map $\tau:A \to A$, and since $[A,x^{p^n}] \ne 1$, the multiplicative order of $\tau$ is a power $p^k$ of $p$ with $k > n$.
So the minimal polynomial $\mu_{\tau}$ of $\tau$ (which I will write as a polynomial in the indeterminate $z$, to avoid double use of $x$) divides $z^{p^k}-1 = (z-1)^{p^k}$, but it does not divide $(z-1)^{p^n}$. So $\mu_{\tau}(z) = (z-1)^m$ for some $m$ with $p^n<m \le p^k$.
So there exists $b \in A$ with $(\tau - 1)^mb= 0$ but $(\tau-1)^{m-1}b \ne 0$. Let $a = (\tau-1)^{m-p^n}b$.
Then $(\tau - 1)^{p^n}a = 0$, but $(\tau - 1)^{p^n-1}a \ne 0$. So $(z-1)^{p^n}$ is the minimal polynomial of $\tau$ in its action on $a$, and hence the $p^n$ elements $\{\tau^i a : 0 \le i < p^n \}$ of $A$ are linearly independent.
Moving back to the group theory, we have proved that the elements $\{ a^{x^i} : 0 \le i < p^n \}$ of $A$ generate a subgroup of $A$ of order $p^{p^n}$, and $(\tau-1)^{p^n} a = 0$ is equivalent to $[a,x^{p^n}]=1$. These are exactly the conditions we need for $\langle a,x \rangle/\langle x^{p^n} \rangle \cong C_p \wr C_{p^n}$.