Show that for every $g \in G$, $g$ lies in the subgroup generated by $P$ and $g^{−1}Pg$.

Question

Let $G$ be a finite group and let $P$ be a Sylow $p$-subgroup of $G$. Suppose that the number of Sylow $p$-subgroups of $G$ is at least $|G : P|$, where $|G : P|$ is the index of $P$ in $G$. Show that for every $g \in G$, $g$ lies in the subgroup generated by $P$ and $g^{−1}Pg$.

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My attempt:

Let $n_p$ denote the number of Sylow $p$-subgroups contained in $G$, then we know $n_p=\frac{|G|}{p^n}$ where $|G|=p^n\times q$ with $\gcd(p,q)=1$.

I have tried to let $G$ act on the left cosets $\{P,a_1P,...,a_{q-1}P\}$ and I get nothing else useful except there is a bijection between stabilizers of $a_iP$ and Sylow $p$-subgroups of $G$.

How am I supposed to prove that $g\in\langle P,g^{-1}Pg\rangle$?

Answer 1

Let $G$ be a group with a Sylow $p$-subgroup, $P$. Then, the number of Sylow $p$-subgroups of $G$, $n_p(G)$ is the index of $\_\_\_\_\_$ in $G$. Given in this case that $n_p(G)\geq [G:P]$, we have that the order of $\_\_\_\_\_$ is at least the order of $\_\_\_\_\_$ so we get that $\_\_\_\_\_=\_\_\_\_\_$.

$g^{-1}Pg$ is a Sylow $p$-subgroup of $G$ and so is also a Sylow $p$-subgroup of $\langle P,g^{-1}Pg\rangle$.

By Sylow's second theorem, there must exist some $h\in \langle P,g^{-1}Pg\rangle$ such that $\_\_\_\_\_=\_\_\_\_\_$. From this we conclude $gh^{-1}\in\_\_\_\_\_$.

Using the fact shown in the first paragraph, we then have $g=fh$ for some $f$ in $\_\_\_\_\_$.

We conclude then that $g\in \langle P, g^{-1}Pg\rangle$.

(Full solution:)

In general, $n_p(G)=[G:N_G(P)]$, where $N_G(P)$ is the normalizer of $P$ in $G$. Since, $n_p(G)\geq [G:P]$ in this case, $[G:N_G(P)]\geq [G:P]$. By Lagrange's theorem, we then have $|N_G(P)|\leq |P|$. Combined with the fact that, in general, $P\leq N_G(P)$, we have $P=N_G(P)$. Given that $P$ and $g^{-1}Pg$ are both Sylow $p$-subgroups of $\langle P, g^{-1}Pg\rangle$, we have that $P$ and $g^{-1}Pg$ are conjugate in $\langle P, g^{-1}Pg\rangle$, i.e. there exists some $h\in \langle P, g^{-1}Pg\rangle$ such that $h^{-1}Ph=g^{-1}Pg$. For this $h$, we then have $gh^{-1}Phg^{-1}=P$, i.e. $gh^{-1}\in N_G(P)$. Since we previously proved $N_G(P)=P$, we have $gh^{-1}\in P$, so for some $f\in P$, $g=fh$. Since $h\in \langle P, g^{-1}Pg\rangle$ and $f\in P\leq \langle P, g^{-1}Pg\rangle$, $g=fh\in \langle P, g^{-1}Pg\rangle$.

Answer 2

Fun thing to observe: a general converse is also true. For Sylow $p$-groups this yields an if and only if so to speak.

Proposition Let $J$ be a subgroup of the group $G$ with the property that for every $g \in G$, $g \in \langle J,J^g \rangle$. Assume that $J \leq H \leq G$, then $N_G(H)=H$.

Proof Let $g \in N_G(H)$. Since $J \leq H$, we also have $J^g \leq H^g=H$. Hence $\langle J, J^g\rangle \leq H$. And the property of $J$ guarantees $g \in H$.

Note The group $J$ with the property above is sometimes called abnormal. Observe that if $P \in Syl_p(G)$, then $N_G(P)$ is abnormal.