Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
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
add comment

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}

share|improve this answer
I got it! Thank you. –  Obaka Feb 5 '13 at 20:27
@Obaka, you're welcome. –  Andreas Caranti Feb 5 '13 at 21:14
add comment

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

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

share|improve this answer
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
add comment

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.