Let $p$ be a prime number and $m, k$ two positive integers. Then ${p^km \choose p^k} \equiv m \pmod p$. I've been trying to demonstrate this lemme all the day. Have you got any suggestion? Thank you in advance.
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2See http://people.brandeis.edu/~igusa/Math101aF07/Math101a_notesA6.pdf. – Dietrich Burde Mar 06 '14 at 22:11
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Here is a group action proof. Let $G$ be a cyclic group of order $p^km$ and let $H$ be the subgroup of $G$ of order $p^k$. Let $H$ act on the set $X$ of subsets of $G$ of size $p^k$. Then since $H$ is a $p$-group, the number of fixed points of $H$ on $X$ is congruent to the size of $X$ modulo $p$. But a subset of size $p^k$ is fixed by $H$ iff it is a coset of $H$. Since there are $m$ cosets of $H$, we conclude that $\binom{p^km}{p^k}=|X|\equiv m\bmod p$.
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How do you prove that if $A \subseteq G$ has size $p^{k}$ and it is fixed by $H$, then $A$ must be a coset? I only managed to prove the other direction, namely, every coset has $p^{k}$ elements and is fixed by $H$. – Diego Mathemagician Aug 09 '22 at 12:16
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1@DiegoMathemagician, the orbits of $H$ acting on $G$ are the cosets and so if $H$ fixes $A$, that is $A$ is invariant under $H$, then $A$ must be a union of cosets. Since $|A|=|H|$, it can only be that $A$ is a single coset. – Benjamin Steinberg Aug 09 '22 at 12:28
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This is a special case of Lucas's theorem.
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Ah good to know; I was about to type out my own answer which is easily superseded by Lucas's theorem! – suvrit Mar 06 '14 at 22:12