This is a documentation of some interesting exercises in Bourbaki.
Let \(p\) be a prime and \(r, t\) non-negative integers, \(C\) is a cyclic group of order \(p^{r+t}\) and \(T\le C\) a subgroup of order \(p^t\). So \(p^r = [C:T] = |C/T|\).
If \(C\) acts on a finite set \(E\), we have
\[ |E| \equiv \left|E^T\right| \mod p^{r+1} = p[C:T].\]
think of \(E^T\) as something that interpolate between \(E\) and \(E^C\), where \(|E|=|E|\) and \(|E|\equiv |E^C|\mod p\).
As an application, we have
\[ \binom{p^{r+t}m}{p^tn} \equiv \binom{p^rm}{n} \mod p^{r+1}. \]
This is an application of orbit formula
\[ |E| = \sum [C:\mathrm{Stab}(e)] \]
we have that \(e\in E^T\) iff \(T\subset \mathrm{Stab}(e)\) iff \([C:\mathrm{Stab}(e)]\) is a factor of \([C:T]\) (Note this iff requires cyclic!). So that otherwise, \(e\not\in E^T\) iff \([C:\mathrm{Stab}(e)]\) is divisible by \(p[C:T]\), thus the result.
Let \([n]\) denote a finite set of size \(n\). Consider the action of \(C\) on \(C\times [m]\) given by acting on the first coordinate. This induces an action of \(C\) on \(E=\binom{C\times [m]}{|T|n}\), the set of subsets of \(C\times [m]\) of size \(p^tn=|T|n\).
The projection map \(\varphi: C\times [m] \to C/T\times [m]\) is \(C\)-equivariant and also \(T\)-invariant (though these facts are not used). We claim that this induces a bijection
\[ \varphi': E^T = \binom{C\times[m]}{|T|n}^T \to \binom{C/T\times [m]}{n}.\]
If a set \(F\in E\) is fixed by \(T\), it is given a \(T\)-action on \(F\). This action is free (all stablizers are trivial) so every orbit is of size \(|T|\), mapping each orbit (there are \(n\) of them) \(T\cdot(s,x)\) to \(Ts\in C/T\) gives the desired bijection. This shows
\[ \left|E^T\right| = \left|\binom{C/T\times [m]}{n}\right| = \binom{p^{r}m}{n}.\]
Applying the previous result, and use \(|E| = \binom{p^{r+t}m}{p^tn}\), we get
\[ \binom{p^{r+t}m}{p^tn} \equiv \binom{p^rm}{n} \mod p^{r+1}. \]
The above proof inspires me to come up with a new proof of the binomial expansion mod \(p\), i.e.
\[\binom{n}{m} \equiv \binom{n_k}{m_k}\cdots \binom{n_0}{m_0} \mod p\]
where \(n = n_kp^k + n_{k-1}p^{k-1} + \cdots + n_0\) and \(m = m_kp^k + m_{k-1}p^{k-1} + \cdots + m_0\) with \(0\le m_i, n_i < p\).
Let \(C_i\) be the cyclic group of order \(p^i\). Consider the set
\[X=\bigsqcup_{i=0}^k C_i \times [n_i].\]
This set is endowed with a natural \(P=C_k\times \dots\times C_0\) action given by multiplying on the first coordinate. The action naturally induces an action on the set \(E=\binom{X}{m}\) of size \(m\) subsets of \(X\).
Now let’s investigate what is \(E^P\). We have that a set \(F\in E^P\) would imply a \(P\) action on \(F\), for which the orbit sizes are computed as
\[|P\cdot (c_i,x)| = [P:\mathrm{Stab}((c_i,x))] = |C_i| = p^i.\]
This together with the uniqueness of \(p\)-adic expansions tells us that the size of \(F\cap (C_i\times [n_i])\) must be \(p^im_i\), consisting of exactly \(m_i\) orbits of size \(p^i\). Collapsing the orbits and mapping to the set \(\prod_{i=0}^k \binom{[n_i]}{m_i}\) gives the bijection
\[ E^P = \binom{X}{m}^P \xrightarrow{1:1} \prod_{i=0}^k \binom{[n_i]}{m_i}.\]
Finally, utilizing the orbit formula, for any \(p\)-group \(|E|\equiv |E^P| \mod p\), we get the desired result
\[\binom{n}{m} \equiv \binom{n_k}{m_k}\cdots \binom{n_0}{m_0} \mod p.\]