What is the cardinality of the set of all subgroups of $S_\alpha$ for infinite $\alpha$? Let’s define $S_\alpha$ as a group of all bijections on a set with cardinality $\alpha$ under composition.
Let’s denote the set of all subgroups of a group $G$ as $Sub(G)$. 

What is the cardinality of $Sub(S_\alpha)$ for $\alpha \geq \aleph_0$?

Currently I only know the following bounds:
$$2^\alpha \leq |Sub(S_\alpha)| \leq 2^{2^\alpha}$$
Proof of the lower bound:
There are known to be $2^\alpha$ non isomorphic groups of order $\alpha$, and each of them is isomorphic to some subgroup of $S_\alpha$ by Cayley theorem.
Proof of the upper bound:
$$Sub(S_\alpha) \subset P(S_\alpha)$$
Thus
$$|Sub(S_\alpha)| \leq |P(S_\alpha)| = 2^{|S_\alpha|} = 2^{2^\alpha}$$
 A: There are $2^{2^\alpha}$ subgroups of $S_\alpha$.
We know that $\alpha=\alpha\cdot 2$ (as a cardinal), so we can use $X=\alpha\times\{0,1\}$ instead of $\alpha$.
For each $A\subseteq \alpha$, consider the following permutation over $X$:
$$f_A((\xi, i))=\begin{cases} (\xi,1-i)&\text{if $\xi\in A$,} \\ (\xi,i)&\text{otherwise}.\end{cases}$$
It is easy to see that $f_Af_B=f_{A\triangle B}$ for all $A,B\subseteq \alpha$ (where $A\triangle B=(A-B)\cup (B-A)$ is the symmetric difference of $A$ and $B$.)
Therefore, the collection of all $G=\{f_A\in S(X)\mid A\subseteq \alpha\}$ forms a subgroup of $S(X)$. Moreover, we can see that the map $A\mapsto f_A$ is an isomorphism between the group $(\mathcal{P}(\alpha),\triangle)$ and $G$, which is isomorphic to the product of $\alpha$ copies of $\mathbb{Z}/2\mathbb{Z}$.
In summary, our $G$ is isomorphic to $(\mathbb{Z}/2\mathbb{Z})^\alpha$. We can see that the dimension of $(\mathbb{Z}/2\mathbb{Z})^\alpha$ is $2^\alpha$, so we can generate $2^{2^\alpha}$ subspaces of $(\mathbb{Z}/2\mathbb{Z})^\alpha$.
