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Let $G$ be a finite group, $p$ a prime dividing the order of $G$, and let $S=\{ P: P \text{ is a Sylow }p \text{-subgroup of }G\}$. Let $x\in G.$ Then $S=S^{x}.$

Proof. Since the elements of $S^{x}$ are Sylow subgroups of $G$, then $S^{x} \subseteq S$ .Define the function $f:S \rightarrow S^{x}$ by $f(P)=P^{x}$. Let $P_{1}, P_{2}$ be elements of $S$.

$f$ is well defined. If $P_{1}=P_{2}$, then $P_{1}^{x}=P_{2}^{x}$ and so $f(P_{1})=f(P_{2})$.

$f$ is one-to-one. If $f(P_{1})=f(P_{2})$, then $P_{1}^{x}=P_{2}^{x}$ and so $P_{1}=P_{2}$.

$f$ is onto. Let $P^{x} \in S^{x}$. Then we have $P \in S$ such that $f(P)=P^{x}$.

Thus $S=S^{x}$

Is the above true?

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To me it seems that you have shown that $S$ and $S^x$ have the same cardinality (same number of elements) by constructing a bijection between $S$ and $S^x$, but this is not enough (and not needed at all for the proof) for the equality $S = S^x$. What you need to show is that every element of $S$ is in $S^x$ and conversely, every element of $S^x$ is in $S$. Then the two sets are equal. – Mikko Korhonen Nov 12 '12 at 17:01
@m.k.: Since the elements of $S^{x}$ are Sylow subgroups of $G$, then $S^{x}\subseteq S$. By the above proof $|S^{x}|=|S|$ and so we must have $S^{x}=S$. – user28083 Nov 12 '12 at 17:09
Another approach is to map $S^x$ back to $S$ by the map taking any $P$ to $P^{x^{-1}}$, which is inverse to the map taking $P$ to $P^x$. – coffeemath Nov 12 '12 at 17:27
Thanks for every one. – user28083 Nov 13 '12 at 13:19

Well, because of the finiteness, if you have a bijection and $S^x\subseteq S$ then it indeed follows that $S^x=S$.

Probably, it's simpler to argue further that $S^{x^{-1}}\subseteq S$, and now using your bijection $f$, we get $S=\big(S^{x^{-1}}\big)^x \subseteq S^x\subseteq S$.

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Thank you very much. – user28083 Nov 13 '12 at 13:19

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