What is multiplicative group $\bar{\mathbb{F}}_p^*$ for algebraic closure $\bar{\mathbb{F}}_p$ of $\mathbb{F}_p$ where $p$ is prime? We know a lot of subgroups of  $\bar{\mathbb{F}}_p^*$:
$\mathbb{Z}/q\mathbb{Z} \subset \bar{\mathbb{F}}_p^*$ if $q \ne p$ is prime. It easy to check using fact that $\mathbb{F}_{p^n} \subset \bar{\mathbb{F}}_p$ and the fact that for every $q \ne p$ prime there exist $n$ such that $p^n - 1$ is divisible by $q$.
Is it true, that $\mathbb{Z}/(q\mathbb{Z})^m\subset \bar{\mathbb{F}}_p^*$? What about $\mathbb{Z}/(q^m\mathbb{Z})\subset \bar{\mathbb{F}}_p^*$? And  what other subgroups does it have?
 A: You can write the algebraic closure of $\mathbb{F}_p$ as
$$ \lim_{\longrightarrow} \mathbb{F}_{p^{n!}} $$
This can be thought of as the union of the $\mathbb{F}_{p^{n!}}$ by gluing them appropriately. The reason to introduce the $n!$ is to ensure that these can be glued.
Then, a finite subgroup $H \subset \overline{\mathbb{F}_p}^*$ is contained in $\mathbb{F}_{p^{n!}}$ for some $n$. Since the unit group of each of these is cyclic, it is easy to sea that $\mathbb{Z}/m\mathbb{Z}$ is a subgroup of $\overline{\mathbb{F}_p}^*$ if and only if $m | p^{n!} - 1$ for some $n$.
This happens if and only if $m$ is coprime to $p$ (remember that $m| p^{\phi(m)} - 1$ if they are coprime).
Observe that this actually characterises all of the finite subgroups.
If you want a more specific description, note that we can take $n$th roots of any $n$ coprime to $p$. Hence, this group is isomorphic to the localisation of the integers away from $p$. That is,
$$ \overline{\mathbb{F}_p}^* \cong \mathbb{Z}_{(p)} / \mathbb{Z}$$
(don't confuse the last bit of notation with $p$-adics)
A: In fact, the group is isomorphic to
$$
\prod_{q\neq p} \mu_{q^{\infty}}
$$
where product runs over all primes not equal to $q$ and $\mu_{q^{\infty}}$ is a $q$-prufer group
$$
\mu_{q^{\infty}} = \mathrm{colim}_{n}\mathbb{Z}/q^{n}\mathbb{Z}.
$$
In other words, it is a smallest abelian group which contains all finite cyclic groups order prime to $p$.
To show this, we'll use the following facts:

*

*The colimit of all finite cyclic groups is isomorphic to
$$
A:=\mathrm{colim}_{n} \mathbb{Z}/n\mathbb{Z} \simeq \prod_{q}\mu_{q^\infty}
$$
by Chinese remainder theorem.


*The algebraic closure of finite field $\mathbb{F}_{p}$ is isomorphic to the colimit of all finite extensions
$$
\mathrm{colim}_{n} \mathbb{F}_{p^n}.
$$
Note that $\mathbb{F}_{p^n} \subseteq \mathbb{F}_{p^m}$ when $n|m$. By taking the unit group, we get
$$
\bar{\mathbb{F}}_p^{\times}  \simeq \mathrm{colim}_{n} \mathbb{F}_{p^n}^{\times} \simeq \mathrm{colim}_{n} \mathbb{Z}/(p^{n} - 1)\mathbb{Z}
$$
which is naturally a subgroup of the above group $A$.


*For any prime $q\neq p$ and $e\geq 1$, there exists $n$ such that $q^{e}$ divides $p^{n} - 1$. This is because of the Euler's theorem: we can simply set $n = \phi(q^{e}) = q^{e-1}(q-1)$.
Now, we are ready to prove our claim. First, for any prime $q\neq p$ and $e\geq 1$, there exists $n$ such that $\mathbb{Z}/q^{e}$ is a subgroup of $\mathbb{Z}/(p^{n}- 1)$. By using Chinese Remainder Theorem, this proves that any finite cyclic group whose order is prime to $p$ is a subgroup of $\bar{\mathbb{F}}_{p}^{\times}$. So we have
$$
\prod_{q\neq p}\mu_{q^{\infty}} \subseteq \bar{\mathbb{F}}_{p}^{\times} \subset A = \prod_{q}\mu_{q^{\infty}}.
$$
The equality follows from the fact that our group can't contain element of order $p$.
