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Question. For $n\geq2$, the following is always an integer. Is it not? $$\frac{(2^n-2)(2^{n-1}-2)\cdots(2^3-2)(2^2-2)}{n!}.$$


2 Answers 2


As $S_n$ also embeds in $GL_{n-1}({\mathbb F}_2)$, you only need to check that $n!$ is not divisible by $2^n$.

  • $\begingroup$ But it never is. $\endgroup$ Mar 24, 2017 at 23:10
  • $\begingroup$ (1) It is possible for $2^{n-1}$ to divide $n!$. (2) $\mathfrak{S}_n$ embeds in $GL_n(\mathbb{F}_2)$ as permutation matrices, but how does it embed in $GL_{n-1}(\mathbb{F}_2)$? (3) At any rate, although I like to approach again, however this answer need more details. I wish it works. $\endgroup$ Mar 24, 2017 at 23:17
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    $\begingroup$ If $V={\mathbb F}_2^n$ with the permutation action of $S_n$ and $W$ is the subspace of $V$ spanned by $(1,1,\ldots,1)$, then $W$ is $S_n$-invariant and thus $S_n$ acts on $V/W$. The almost simplicity of $S_n$ guarantees that this action is faithful for all $n>4$, and you can check the smaller cases. $\endgroup$ Mar 24, 2017 at 23:26
  • $\begingroup$ Incidentally, there's a subspace $X$ of $V$ consisting of all vectors whose entries sum to zero that is also invariant under $S_{n}$. If $n$ is even, then $X$ contains $W$ and this gives an embedding of $S_{n}$ into $GL_{n-2}(\mathbb{F}_{2})$ (provided $n > 4$) and as a corollary, one gets that $\frac{(2^{n} - 4)(2^{n-1}-4) \cdots (2^{3}-4)}{n!}$ is an integer for $n$ even and $n \geq 4$. $\endgroup$ Mar 25, 2017 at 0:34
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    $\begingroup$ Just to explain the last part of John's answer: the group-theoretic argument shows that the quotient $Q=(2^n-2) \cdots (2^2-2)/n!$ is an odd integer times $2^m$ for some $m \in \mathbb{Z}$. Since $n!$ is not divisible by $2^n$, we have $v_2(n!) \leq n-1$ so that $v_2(Q) \geq 0$. $\endgroup$ Mar 25, 2017 at 12:30

Here is a direct number theoretic proof. First of all note that $$(2^n-2)(2^{n-1}-2)\cdots(2^2-2)=2^{n-1}(2^{n-1}-1)(2^{n-2}-1)\cdots(2^1-1).$$ Now for any prime $p$ we have, by Legendre's formula and by Fermat's little theorem, $$v_p(n!)\leq\left\lfloor\frac{n-1}{p-1}\right\rfloor\leq v_p\bigl(2^{n-1}(2^{n-1}-1)(2^{n-2}-1)\cdots(2^1-1)\bigr),$$ where for the second inequality we argue separately for $p=2$ and for $p\geq 3$. The result follows.

P.S. The above proof was inspired by Cherng-tiao Perng's deleted response (which was not entirely correct).

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    $\begingroup$ Indeed. I should have seen this and mentioned to Cherng-tiao how to fix it. Gerhard "Tenders His Most Sincere Apologies" Paseman, 2017.03.24. $\endgroup$ Mar 25, 2017 at 1:16

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