Is there a geometrical interpretation of this equality $2\cdot 4\cdot 6\cdot\ldots\cdot(2n)=2^nn!$?

Question

$$2\cdot 4\cdot 6\cdot\ldots\cdot(2n)=2^nn!$$

How it can be seen in a plane?

I have found many proofs with by induction but I wish to understand it geometrically.

Answer 1

I've actually been meaning to think about this exact question for a while, in a specific context:

Theorem: The $n$-dimensional cube $Q_n$ has exactly $2^n \cdot n!$ symmetries, including reflections.

I've always seen it listed that the size $\left| \operatorname{Sym}(Q_n)\right|$ of the symmetry group of $Q_n$ is $2^n \cdot n!$, and only recently did I learn this happens to be equal to the double factorial $(2n)!! = 2 \cdot 4 \cdot \ldots \cdot 2n$. But shamefully, I've never known why either should count symmetries of the cube!

We'll look first at why $\left| \operatorname{Sym}(Q_n)\right| = 2^n \cdot n!$. We'll pick a particular vertex $V$ of $Q_n$, say the front-upper-left one in the following picture.

Now, the orbit-stabilizer theorem tells us that

$$\left| \operatorname{Sym}(Q_n)\right| = (\text{number of places $V$ can go}) \cdot (\text{number of symmetries fixing }V),$$

where it's apparent that $V$ can get sent to any of the other $2^n$ vertices. But counting symmetries that fix $V$ can be a bit tricky. More pictures are in order.

On the left we see vertex $V$ in red, and some rotations implied by the green arrows. On the right, there's a hyperplane (implied in green) through which we can reflect, to obtain another symmetry.

But what the two pictures really have in common is the blue triangle: it's the key! To get the triangle, take the vertices vertices that are just one step away from $V$ and connect them (see also, vertex figure). A moment's thought should convince you of two things:

• That for an $n$-dimensional cube, you won't necessarily get a $2$-dimensional triangle, rather an $(n - 1)$-dimensional simplex.

• Further, symmetries fixing $V$ are exactly symmetries of this simplex; symmetries fixing $V$ are in bijection with symmetries of a uniform simplex.

However, counting symmetries of the $(n-1)$-simplex is quite tractable, and there are $n!$ of these, hence $\left| \operatorname{Sym}(Q_n)\right| = 2^n \cdot n!$.

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Now it remains to show that $\left| \operatorname{Sym}(Q_n)\right| = 2 \cdot 4 \cdot \ldots \cdot 2n = (2n)!!$. Our approach will appeal to the fact that the cube $Q_n$ is a regular polytope, which means that its symmetry group acts transitively on its set of flags.

A flag, you say, more terminology? Yes, but it's quite manageable. A flag is simply a nested set of faces of the cube, one for each dimension. For example, in the $3$-dimensional cube $Q_3$, a flag looks like

$$\text{vertex $\subset$ edge $\subset 2$-dimensional face}.$$

Now, the definition of regularity just says that we can move any such chain of nested faces to any other chain, using one of the symmetries of the cube.

Put another way, symmetries of the cube are in bijection with flags of the cube! Now, those are quite easy to count, since each face of the cube $Q_n$ is some cube $Q_k$ of dimension $k$, which has exactly $2k$ faces of dimension $k-1$.

To start specifying a flag in the $n$-dimensional cube $Q_n$, we have $2n$ faces of maximal dimension $n-1$ to choose from (e.g., $6$ faces of dimension $2$ in $Q_3$). For our next step down, there are $2(n - 1)$ faces of dimension $n - 2$, since we're only allowed to choose a maximal face of the face we've already chosen (e.g., having chosen one of the six square faces in $Q_3$, we choose one of its $4$ edges). We continue like this until we've chosen one of the $4$ vertices of a $2$-dimensional face, at which point we choose one of its $2$ vertices. Now we're done, having made $2n \cdot 2(n - 1) \cdot \ldots 4 \cdot 2$ choices and completely determined a flag!

Thus $$2^n \cdot n! = \left| \operatorname{Sym}(Q_n)\right| = 2 \cdot 4 \cdot \ldots \cdot 2n.$$