1
$\begingroup$

Let $P \in Syl_p(G)$ and assume N is normal in G. Use conjugacy part of sylow's theorem to prove that $P \cap N$ is a sylow p-subgroup of N. Deduce that PN/N is a sylow p-subgroup of G/N.

How do we know that N will contain p-subgroup of G of same order as P ? I don't understand this question could somebody explain more.

$\endgroup$
  • $\begingroup$ The first part does not imply that $|P \cap N| = |P|$. It only implies that $[N:P \cap N]$ is not divisible by $p$. $\endgroup$ – Geoff Robinson Oct 4 '15 at 20:27
0
$\begingroup$

Write $|G|=p^r m$ with $p \nmid m$ for some $r \geq 1$. By assumption, $|P| = p^r$. Since $N \triangleleft G$, it follows that $P \cap N \triangleleft P$ and so by Lagrange's Theorem, $|P \cap N| = p^s$ for some $1 \leq s \leq r$. Now, we have that $$|PN| = |P| \cdot |N|/|P \cap N| \quad \implies \quad [N : P \cap N] = [PN : P].$$ Since $[G:P] = [G : PN] [PN : P]$, we must have that $[PN : P] = [N : P \cap N]$ divides $[G:P] = m$. Can you take it from here?

For the second part, use the Second Isomorphism Theorem to get $PN/N \cong P/(P \cap N)$. Show that $PN/N$ is a $p$-subgroup of $G/N$ and use the above result.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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