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First of all, let me describe the problem that I try to solve:

Let $p$ be a prime, $G$ a finite group, $P$ a $p$-Sylow subgroup of $G$, and $L$ a set of all elements of $G$ which its order is coprime to $p$. Prove that if $K$ is a normal subgroup of $G$ with $\lvert K \rvert$ is coprime to $p$ and $G = PK$ then $K = L$.

I already have proved that $K \subset L$. (For any element $k$ of $K$, consider $\langle k \rangle$ and use Lagrange's theorem and the hypothesis that $\lvert K \rvert$ is coprime to $p$.)

Then I try to show $K \supset L$, but I cannot. Could you help me? The following is what I tried:

Let $z$ be a element of $L$. From $L \subset G = PK$, it can be written of the form $z = xy$ for some $x \in P$ and $y \in K$. Put $q$ be the highest power of $p$ dividing $\lvert G \rvert$. Since $K$ is normal and $P$ is $p$-Sylow subgroup, $$ z^q = (xy)^q = x^q k = k \in K $$ for some $k \in K$. Hence, at least, $z^q \in K$. But from here, how can I show $z \in K$? If my English is bad or mathematical description is unclear then tell me; I'll fix it. Thank you.

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Hint: Let $m$ be the order of $k$. Since $m$ is coprime to $q$ there is some integer $n$ such that $nq$ is congruent to $1$ mod $m$. –  Tobias Kildetoft Feb 5 '13 at 20:18
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2 Answers

up vote 3 down vote accepted

If I understand your question correctly, you are nearly there. Let $z \in L$, $n = \lvert P \rvert$, $m = \lvert G : P \rvert$. Then $z^n \in K$, basically as in your proof. But you have also $z^m = 1$, as $z \in L$. Since $(m, n) = 1$, there are $u, v$ such that $m u + n v = 1$. Thus \begin{equation} z = z^1 = z^{m u + n v} = (z^{m})^u (z^{n})^v \in K. \end{equation}

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I got it! Thank you. –  Obaka Feb 5 '13 at 20:27
    
@Obaka, you're welcome. –  Andreas Caranti Feb 5 '13 at 21:14
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I think you forgot that the order of $\,z\,$ is coprime with $\,p\,$ and, thus, with $\,q=p^n\,$, too, so if $\,t=\mathcal Ord(z)\,$ , then there exist $\,m,n\in\Bbb Z\,$ s.t. $\,mt+nq=1\,$ , but then

$$z=z^1=z^{mt+nq}=(z^t)^n(z^q)^n=k^n\in K$$

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I think the last equation is $$ \cdots = (z^t)^m(z^q)^n = k^n \in K. $$ Anyway, I got an idea. Thank you! –  Obaka Feb 5 '13 at 20:48
    
Indeed, a little typo that doesn't change anyhting essential. –  DonAntonio Feb 5 '13 at 21:00
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