Write $A$ for the group of continuous automorphisms of $\mathbb{F}_{p}[[t]]$ and for each $n \geq 1$ let $J_{n}$ be the kernel of the homomorphism from $A$ to the automorphism group of $\mathbb{F}_{p}[[t]]/(t^{n+1})$, where $(t^{n+1})$ is the ideal generated by $t^{n+1}$.
For each $n \geq 1$ let $e_{n} \in J_{n}$ be defined by $e_{n}(t) = t + t^{n+1}$. Prove that every element of $J_{n}$ is congruent modulo $J_{n+1}$ to a power of $e_{n}$.
Now, I've borrowed amount of books relating to p-groups in general from the library, and I've managed to find a book that has the solution to this particular problem:
$J = J_{1} > J_{2} > J_{3} > \dots$ , and each section in this series has order $p$; and hence $J_{n} = \big \langle e_{{n}} \rangle J_{n+1}$ for all $n \geq 1$.
That each section has order $p$ is clear, but the I don't understand why the statement after "hence" follows. Could someone please explain why this is the case? Many thanks.
Since $|J_n / J_{n+1} |$ is prime, there is simply no room for any subgroup strictly between $J_n$ and $J_{n+1}$ :
$J_{n+1} \subsetneq \langle e_n \rangle J_{n+1} \subseteq J_n$, so $\langle e_n \rangle J_{n+1} / J_{n+1}$ is a nontrivial subgroup of $J_n/J_{n+1}$.
By Lagrange's theorem, its size has to divide $p$. Since it can't be $1$ it has to be $p$,
and so $\langle e_n \rangle J_{n+1} / J_{n+1} = J_n / J_{n+1}$, which implies $\langle e_n \rangle J_{n+1} = J_n$.