Let $p$ be a prime and let $H$ be a subgroup of an infinite abelian group $\Gamma$, and suppose we have a chain of elements in $H$ $$ h_0, h_1, \ldots $$ such that $h_{n+1}^p = h_n$. Does it follow from here that $H$ must be trivial? If so, how?
Context:
In the problem where this arises, I have that $\Gamma$ is the maximal abelian Galois group of a characteristic $p$ local field $K$, and $H$ is the subgroup of $\Gamma$ fixing a known subfield $L$ of the maximal abelian Galois extension of $K$, denoted $K_{ab}$. I would like to show that $L = K_{ab}$, which I've been advised to do by assuming that $L = (K_{ab})^H$ and then showing that $H$ must be trivial, which I've been told can be deduced from the statement above. Here's the chain of reasoning that was provided, as a rough guide:
- For any $g \in \Gamma$ whose image in $\Gamma/H$ is of finite order $m$, we must have $m$ divides $q - 1$, for $q$ a power of $p$. This has to do with the structure of $\Gamma/H$, which is $$ \Gamma/H = \hat{\mathbb Z} \times \mathbb F_{q}^\times \times U$$ where $U$ is the group of principal units of the valuation ring of $K$.
- Hence if $g \in \Gamma$ such that $g^p \in H$, then $g \in H$.
- The field L contains all cyclic extensions of $K$ of degree $p$, and $K_{ab}$ appears as the union of finite field extensions $E_i$ of $L$, such that $[E_i:L] = p^{m_i}$ for $m_i \in \mathbb N$.
- From 3, we see that $H\subset (\Gamma)^p$. (I'm a little fuzzy on the reasoning behind this `fact' as of now)
- For any $h \in H$, we can form a chain of elements of the form $$ h_0, h_1, \ldots \in H $$ such that $h_0 = h$ and $h_{n+1}^p = h_n$.
- Deduce from 5 that $H = \{e\}$.