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Algebra

Wielandt's proof of Sylow's Theorem

Wieldandt's proof of Sylow's Theorem A brief preparatory combinatorial discussion: [ \binom{n}{k} = \dfrac{n!}{k!(n-k)!} = \dfrac{n (n-1) \ldots (n-j) \ldots (n-k+1)}{ 1 \ldots j \ldots (k-1) \cdot k } = \dfrac{\prod_ {j=1}^k (n-j+1) }{ \prod_ {j=1}^k j } ] and therefore [ \binom{n}{k} = \dfrac{n}{k} \cdot \dfrac{n-1}{1} \ldots \dfrac{n-j}{j} \ldots \dfrac{n-(k-1)}{k-1} = \dfrac{n}{k} \prod_{j=1}^{k-1} \dfrac{n-j}{j} ] Let $p$ be a prime number. Now, if $n$ is an integer divisible by a power of $p$, say $p^\alpha$, then $n=p^\alpha m$.

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