TITLE: A result from Peter McMullen's thesis
QUESTION [7 upvotes]: The classical definition of regular polytopes is recursive. It says that a polytope is regular if its facets and vertex figures (both smaller-dimensional polytopes) are regular.
The modern definition goes as follows. Let $P\subseteq\mathbb{R}^n$ be a polytope centered at the origin and let $\mathrm{Aut}(P)\leq O(n)$ be its automorphism group. We say that $P$ is regular if its automorphism group acts transitively on maximal flags of faces.
However, there are many equivalent definitions of regularity. Let's say that a polytope is $d$-regular if its automorphism group is transitive on $d$-dimensional faces. The following theorem is stated in several places (for example in McMullen and Schulte's "Abstract Regular Polytopes", pages 9-10):
Theorem: Let $P$ be an $n$-dimensional polytope. If $P$ is $d$-regular for all $0\leq d\leq n-1$ then $P$ is regular.
All statements of this theorem I've seen refer to Peter McMullen's 1968 thesis from the University of Birmingham, which I don't have access to.
So here's my question: Does anyone know where I can find a proof of this theorem or how to gain access to Peter McMullen's thesis?
Bonus Problem: How dependent/independent are the notions of $d$-regularity for different $d$?
Thanks.

REPLY [8 votes]: This is theorem 4C6 of Peter McMullen's thesis, "On the Combinatorial Structure of Convex Polytopes", on page 73:

4C6. Theorem. A $d$-polytope $P$ is regular if and only if for each $j = 0, \dots, d-1$, its symmetry group $G(P)$ is transitive on the $j$-faces of $P$.
The condition is clearly necessary. For the converse, we deduce that
  the vertices of $P$ lie on a $(d-1)$-sphere, $S^{d-1}$, and hence that
  the vertices of any $j$-face $F^j$ $(j=1,\dots,d)$ lie on the
  $(j-1)$-sphere $S^{d-1} \cap \operatorname{aff} F^j$. If we assume
  inductively that the facets of $F^j$ are regular and congruent, then
  by theorem 4C4, we deduce that $F^j$ is itself regular. In particular,
  for $j = d$, we deduce that $P$ is regular. This completes the proof
  of theorem 4C6.

Theorem 4C4 says

4C4. Theorem. A $d$-polytope $P$ ($d \geq 3$) is regular if and only
  if its vertices lie on a sphere, and its facets are regular and
  combinatorially equivalent.
That the condition is necessary is clear, since the transforms of a
  point under a finite group of congruent transformations lie on a
  sphere. We prove that the condition is sufficient by induction on the
  dimension $d$. Since two adjacent facets share a $(d-2)$-face, using
  proposition 1C16 we deduce that any two facets of $P$ are congruent.
  In particular, the edges of $P$ have the same length. This implies
  that the mid-points of the edges of $P$ through any given vertex lie
  on a $(d-2)$-sphere. But these points are the vertices of a
  vertex-figure of $P$ (cf. proposition 1B4); its facets are congruent,
  and so by the induction hypothesis, it is regular. (Notice that this
  holds for the case $d=2$ as well.) In 3 dimensions, this condition
  also implies that the vertex-figures, being regular polygons with the
  same edge-length inscribed in circles of the same radius, are
  congruent. Hence, by theorems 4C2 and 4C3, $P$ is regular, which
  completes the proof of theorem 4C4.

The mentioned results are

1C16: Any two $j$-faces $F^j$ and $G^j$ of a polytope $P$ can be joined by a chain
$$F^j = F_0^j, F_1^j, \dots, F_m^j = G^j$$
of $j$-faces of $P$, such that for $k = 1, \dots, m$, $F_{k-1}^j$ and $F_k^j$ are adjacent.
1B4: $H \cap P \approx P / F$, where $F$ is a vertex of a polytope $P$, and $H$ is a hyperplane strictly separating $F$ from the remaining vertices $P \setminus F$ of $P$.
4C2: A $d$-polytope $P$ ($d \geq 3$) is regular if and only if its facets are regular and combinatorially equivalent, and its vertex-figures are combinatorially regular and combinatorially equivalent.
4C3: A $d$-polytope $P$ ($d \geq 4$) is regular if and only if its facets are regular and its vertex-figures are combinatorially regular.