Let $A$ be an abelian group of order $2^{100}$. Prove that $A$ is not a subgroup of $S_n$ for $n<200$.

Question

Let $A$ be an abelian group of order $2^{100}$. Prove that $A$ is not a subgroup of $S_n$ for $n<200$.

I was thinking of showing that there is no subgroup of order $2^{200}$ by using an argument using lagrange to solve this question by I can't find a way to verify how many $2$s are there in $200!$ and also this is a stronger claim so it seems unprobable. Does anybody have some other ideas?

Edit: I realized we are dealine with $2^{100}$ I made a mistake the question is about order $2^{100}$

Answer 1

You can verify that $2^{101}$ divides $104!$, so looking at just group orders won't get you all the way there.

Instead, look at the possible abelian subgroups of $S_n$. You'll find that such subgroups have a pretty restrictive constraint.

Answer 2

We can prove a very general result: Let $p$ be a prime number, if $S_m$ contains an abelian subgroup $G$ of order $p^n$, then $m\ge np$.

This can be done by induction on $n$. If $n=1$, since $|S_m|=m!$ and $p\not\mid m!$ unless $m\ge p$. In general, if $G\le S_m$, that is $G$ acts on $\{1, 2, \cdots, m\}$ faithfully. Because $G$ is abelian, for each orbit $\mathcal O$ of the action, the stabilizer subgroup $G_x$ with respect to $x\in\mathcal O$ doesn't depend on $x$. If the action is transitive, then for each $x\in \{1, \cdots, m\}$, $G_x$ must be trivial due to faithfullness. Hence $m= p^n\ge np$. Otherwise, since the action is faithful, at least one orbit has more than $1$ element, say $\mathcal O$ and pick $x\in\mathcal O$, we have $G_x$ has cardinality $p^l$ with $0<l<n$. Then $|\mathcal O|=|G|/|G_x|=p^{n-l}$. Because of faithfulness, $G_x$ acts faithfully on $\{1, \cdots, m\}\setminus\mathcal O$, hence $m-|\mathcal O|\ge lp$ by induction hypothesis again. Thus $m\ge p^{n-l}+lp\ge(n-l)p+lp=np$.

Edit:

We can further show whenever the lower bound $m=np$ is achieved, $$ G\simeq \begin{cases}(\mathbb Z/p\mathbb Z)^n & \text{ if } p\not=2 \\ (\mathbb Z/2\mathbb Z)^{n_1}\times(\mathbb Z/4\mathbb Z)^{n_2} & \text{ if } p=2\end{cases}$$

In the above argument, in case of $G$ acts transitively on $\{1, \cdots, m\}$, from $m=p^n=np$ we can deduce $n=1$ or $p=n=2$, because when $n\ge 2$, $$p^n=(1+(p-1))^n=\sum_{i=0}^n {n\choose i} (p-1)^i=1+n(p-1)+\cdots\ge n(p-1)+n=np$$

where equality is achieved iff ${n \choose i}(p-1)^i=1$ for all $i=2, \cdots, n$. So $p=2, n\le 2$.

And in the second case of the induction, for the equality to hold, we must have $p^{n-l}=(n-l)p$. When $p>2$, we have $n-l=1$, hence $|\mathcal O|=p$, in other words, any nontrivial orbit $\mathcal O$ (which are all the orbits when the lower bound is achieved) must have cardinality $p$, and since the only $p$-subgroup of $S_p$ is isomorphic to $\mathbb Z/p\mathbb Z$, we must have $G$ is a subgroup of $(\mathbb Z/p\mathbb Z)^n$, but they have the same cardinality, so $G\simeq(\mathbb Z/p\mathbb Z)^n$. We can similarly handle the case $p=2$ by noting that any maximal abelian $2$-subgroup of $S_4$ is isomorphic to $\mathbb Z/4\mathbb Z$.