TL;DR
Is the entire Clifford hierarchy (as opposed to any one level), a group?
Background.
The Clifford hierarchy (on $n$ qubits), is a collection of nested subsets $\mathcal C^{(1)} \subset \mathcal C^{(2)} \subset \mathcal C^{(3)} \subset \cdots \subset \mathbf U(2^n)$ of the unitary operators on $n$ qubits, defined recursively as follows:
$ \mathcal C^{(1)} $ is the Pauli group on $n$ qubits, i.e., the set of ($n$-fold tensor products of) Pauli operators with real or imaginary scalar factors: tensor products of $\{ \mathbf 1, X, Y, Z \}$ with a scalar factor of $\pm 1$ or $\pm i$.
For $k > 1$, we recursively define $\mathcal C^{(k)}$ as
$$ \mathcal C^{(k)} = \Bigl\{ U \in \mathbf U(2^n) \mathrel{\Big\vert} \forall P \in \mathcal C^{(1)} : U P U^\dagger \in \mathcal C^{(k-1)} \Bigr\}$$ — that is, the set of unitaries that, when used to conjugate a Pauli operator, yields an operator at one level lower in the hierarchy.
So, for example, the Clifford group is the second level of the Clifford hierarchy, $\mathcal C^{(2)}$, as any $U \in \mathcal C^{(2)}$ maps a Pauli operator to another Pauli operator (i.e., an element of $\mathcal C^{(1)}$) by conjugation.
It is well known that $\mathcal C^{(1)}$ and $\mathcal C^{(2)}$ are groups: it is not difficult to show that they are closed under multiplication. It is also well-known that the other levels of the Clifford hierarchy are not groups. For instance: for any $k \geqslant 0$, a $k$-controlled NOT operation $$ \mathrm{C}^k \mathrm X \;=\; \mathbf 1^{\otimes k+1} +\, \lvert 1 \rangle\!\!\;\langle 1 \rvert^{\otimes k} \otimes \,(X - \mathbf 1) $$ is a member of $\mathcal C^{(k+1)} \setminus \mathcal C^{(k)}$, which we can demonstrate by relating these operations to multiply-controlled Z operations $\mathrm{C}^k \mathrm Z = \mathrm{diag}(+1,\ldots,+1,-1)$, and considering their relationships to what we now call phase polynomials. However, using a circuit consisting of 3 such gates, we can easily (using standard computation-uncomputation techniques) produce a circuit which can simulate a $(2k-1)$-controlled NOT operation, which for $k > 1$ does not live in $\mathcal C^{(k+1)}$. (The case $k=2$ amounts to the fact that the TOFFOLI gate, together with preparation of bits in the 1 state, is universal for classical computation.)
However, just because the higher levels of the Clifford hierarchy are not groups, does not mean that the entire Clifford hierarchy is not a group. After all: the proof I give just above that $\mathcal C^{(k)}$ is not a group, establishes this by proving that I can make a product which escapes to some higher level of the Clifford hierarchy.
So: if you multiply two elements of the Clifford hierarchy, do you always get another element of the Clifford hierarchy?