My question is about a short combinatorial discussion in Herstein's Topics in Algebra, on page 92, preceding one of his three proofs of Sylow's Theorem. I will cite the important part:

If $p^r | m$ but $p^{r+1} \nmid m$, consider $$\binom {p^\alpha m}{p^{\alpha}} = \frac{(p^{\alpha}m)!}{(p^{\alpha})!(p^{\alpha}m-p^{\alpha})!} = \frac {p^{\alpha}m(p^{\alpha}m-1) \dots (p^{\alpha}m-i) \dots (p^{\alpha}m-p^{\alpha} + 1)} {p^{\alpha}(p^{\alpha}-1) \dots (p^{\alpha}-i) \dots (p^{\alpha}-p^{\alpha} + 1)}.$$ The question is, What power of p divides $\binom {p^{\alpha} m}{p^{\alpha}}$? Looking at this number, written out as we have written it out, one can see that except for the term $m$ in the numerator, the powers of $p$ dividing $(p^{\alpha}m-i$$)$ is the same as that dividing $p^{\alpha}-i$, so all powers of $p$ cancel out except the power which divides $m$. Thus $p^r | \binom {p^{\alpha} m}{p^{\alpha}}$ but $p^{r+1} \nmid \binom {p^{\alpha} m}{p^{\alpha}}$.

My understanding is as follows. $\binom {p^\alpha m}{p^{\alpha}} = m *\frac{\binom {p^\alpha m}{p^{\alpha}}}{m}$, consider the unique factorization of both factors. The right will not contain any nonzero powers of $p$, and so the power of $p$ dividing our binomial is the largest power in the unique factorization of $m$. But there are two gaps in my understanding: why is the second factor neccesarily an integer, and why does the second factor not contain powers of $p$? Herstein explains this second question, but I don't get his argument.

  • $\begingroup$ I don't see why $\frac1m\binom{p^\alpha m}{p^\alpha}$ need be an integer. Anyway, Herstein neither asserts this nor needs this. $\endgroup$ – Lord Shark the Unknown Mar 24 at 18:11

Kummer's Theorem, which is proven in this answer, says that the number of factors of $p$ in $\binom{n}{k}$ is the number of carries when adding $k$ to $n-k$ in base-$p$.

If $p^r\mid m$ and $p^{r+1}\nmid m$, then adding $1$ to $m-1$ will have exactly $r$ base-$p$ carries. Likewise, adding $p^\alpha$ to $(m-1)p^\alpha$ will have exactly $r$ base-$p$ carries. Therefore, there will be $r$ factors of $p$ in $\binom{mp^\alpha}{p^\alpha}$.

  • $\begingroup$ Thank you for your answer, but isn't there a simpler way to see it? Herstein's book has been pretty self-contained and nothing about carries, p-adic numbers, Kummer's theorem or anything like that has been covered. $\endgroup$ – Steven Wagter Mar 25 at 9:21
  • $\begingroup$ The base-$p$ carries refer to the base-$p$ representation of numbers; not to $p$-adic numbers. The rest of this is detailed in the answer cited above, where it is noted that the number of base-$p$ carries when adding $k$ to $n-k$ equals $\frac{\sigma_p(k)+\sigma_p(n-k)-\sigma_p(n)}{p-1}$ where $\sigma_p(n)$ is the sum of the digits in the base-$p$ representation of $n$. $\endgroup$ – robjohn Mar 25 at 9:32

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