Let $G$ be a group $p$, $q$ primes $p\neq q$, $|G|=p^{100}q$.

1.- Assume that $G$ has two different $p$-Sylow subgroups and intersection for each pair of $p$-Sylow subgroups is trivial. Show that there is a normal $p$-Sylow subgroup in $G$. (Probably this is a teacher mistake and ask for a $q$-Sylow subgroup)

2.- If that each $p$-Sylow subgroup of $G$ is commutative then $G$ is not simple.

Attempt: I tried 1.- and 2.- using Sylow theorem and counting (I think 1.- is a help for 2.-), I look cases when $p<q$, $p>q$ but I don't find a normal subgroup.

My attempt for 2.- is trying to prove for each pair of $p$-Sylow subgroup that the intersection is trivial and use 1.- to get no simplicity with a $q$-Sylow.

Edit: Im offering bounty for who proves the existence of a group of orden $p^{100}q$ where all pairs of $p$ Sylow subgroups intersects trivially, i.e. a non abelian group of order $p^{100}$ which is isomorphic to a semidirect product of one of his $p$ Sylow subgroups and the unique $q$ Sylow (normal) subgroup.

  • 2
    $\begingroup$ In (1) it must be, I think: "Show that there is a normal q-Sylow subgroup...". No Sylow $\;p$-subgroup can be normal as there are more than one of them... $\endgroup$ – DonAntonio Apr 18 '14 at 22:28
  • $\begingroup$ My teacher puts a $p$ in the exam, I also think it must be $q$. I think he is wrong. But in 2.- we probably need to find a $q$- Sylow using 1.- $\endgroup$ – Gaston Burrull Apr 18 '14 at 23:05

Just proffering an alternative solution to part 2 (not needing the normal $p$-complement theorem that Jack Schmidt described).

You know how to do it, if the Sylow $p$-subgroups intersect trivially. What if they don't? Let $x\neq1$ be an element in the intersection of at least two Sylow $p$-subgroups, say $P_1$ and $P_2$. Consider the centralizer of $x$ $$ C_G(x)=\{a\in G\mid ax=xa\}. $$ It is known (or easy to see) that $C_G(x)\le G$. Because the abelian subgroups $P_1$ and $P_2$ are both contained in $C_G(x)$, we know that $C_G(x)$ has more than $p^{100}$ elements. Therefore Lagrange's theorem forces us to have $C_G(x)=G$. Therefore $x$ is in the center of $G$, and thus the center forms a non-trivial normal subgroup of $G$.

  • $\begingroup$ Ty, did you mean $<x>$ is the non trivial normal subgroup of $G$? $\endgroup$ – Gaston Burrull Apr 26 '14 at 20:05
  • $\begingroup$ @Gastón: Either that or the center. The center of a group is always a normal subgroup. The question is whether it is trivial or not. Any subgroup of the center is also normal, so $\langle x\rangle$ works fine also. $\endgroup$ – Jyrki Lahtonen Apr 27 '14 at 4:45
  • $\begingroup$ Ty(13 more to go) $\endgroup$ – Gaston Burrull Apr 27 '14 at 23:16

Here are some examples. They only require matrix multiplication and the existence of finite fields to understand. Below them I also indicate why we need some sort of divisibility condition, and how to solve part 2.

Example: Let $p^a$ be a prime power dividing $q^b-1$, then there is a finite group $G$ of order $|G|=p^a q^b$ with $q^b$ abelian Sylow $p$-subgroups, each pair of which intersects trivially.

Proof: Let $F$ be a field with $q^b$ elements, and let $\alpha$ be an element of $F$ with multiplicative order $p^a$. Define $$G=\left\{ \begin{bmatrix} \alpha^i & \beta \\ 0 & 1 \end{bmatrix} : i \in \mathbb{Z}/p^a\mathbb{Z}, \beta \in F \right\} \leq \operatorname{GL}(2,F)$$ $$P = \left\{ \begin{bmatrix} \alpha^i & 0 \\ 0 & 1 \end{bmatrix} : i \in \mathbb{Z}/p^a\mathbb{Z} \right\} \qquad Q = \left\{ \begin{bmatrix} 1 & \beta \\ 0 & 1 \end{bmatrix} : \beta \in F \right\}$$ Then $P$ is a Sylow $p$-subgroup, $Q$ is the unique Sylow $q$-subgroup. Let $x=\begin{bmatrix} \alpha^k & 0 \\ 0 & 1 \end{bmatrix} \in P \cap P^{g^{-1}}$ for $g=\begin{bmatrix} \alpha^j & \beta \\ 0 & 1 \end{bmatrix}$. Hence $xg =gy$ for some $y=\begin{bmatrix}\alpha^i & 0 \\ 0 & 1 \end{bmatrix} \in P^g \cap P$. However, direct calculation gives that $xg = \begin{bmatrix} \alpha ^{k+j} & \alpha^k \beta \\ 0 & 1 \end{bmatrix}$ while $gy=\begin{bmatrix} \alpha^{i+j} & \beta \\ 0 & 1 \end{bmatrix}$. Hence $\beta = 0$ so that $g \in P$ and $P^g=P$, or $\alpha^k=1$ so that $x=1$. Hence, if $P^g \neq P$, we get $x=1$ is the only element of $P^g \cap P = \{1\}$. More strongly we get that $P$ (and its conjugates) are the only $p$-local subgroups of $G$. Also $P$ is cyclic, and $Q$ is abelian. $\square$

However, the intersection is not forced to be trivial just by $P$ being abelian.

Example: Let $p^a$ be a prime power dividing $q^b-1$ and let $c$ be a non-negative integer. Then there is a finite group $H$ of order $|H|=p^{a+c} q^b$ with $q^b$ abelian Sylow $p$-subgroups, each pair of which intersects in a normal subgroup of order $p^c$.

Proof: Take $G$ form the previous example, and define $H=G \times C_{p^c}$. $\square$

If you don't have any kind of division condition, then there may not be any $G$ with more than one Sylow $p$-subgroup.

I assume familiarity with some early ideas of $p$-subgroups of a finite group, including Sylow's theorems, Burnside's basis theorem, Burnside's fusion theorem, and Frobenius's p-nilpotence criterion. Each of these is fairly directly related to normal $p$-subgroups of a finite group, and each has self-contained proofs that are only a total of a few pages. All could be found in Gorenstein's textbook for a course in Finite Groups.

Proposition: Suppose $G$ is a finite group of order $p^a q^b$ with $p,q$ distinct primes, $G$ has more than one Sylow $p$-subgroup, and every pair of distinct Sylow $p$-subgroups intersect trivially. Then $p$ divides $q^c-1$ for some $1 \leq c \leq b$.

Proof: Let $G$ be such a group. By Frobenius's normal $q$-complement theorem, if $N_G(Q)/C_G(Q)$ is a $q$-group for each $q$-subgroup $Q$, then $G$ has a normal subgroup $P$ of index $q^b$. However, $|G|=p^a q^b$, so $|P|=p^b$ is a Sylow $p$-subgroup. Since all Sylow $p$-subgroups are conjugate, and this one is normal, such a $G$ would have a unique Sylow $p$-subgroup, contrary to our hypothesis. Hence there is some $q$-subgroup $Q$ such that $p$ divides $[N_G(Q):C_G(Q)]$. Let $x \in N_G(Q)$ have order $p$ in the quotient $N_G(Q)/C_G(Q)$. Then $x$ normalizes $Q$ and its Frattini subgroup $\Phi(Q)$, so $x$ induces an automorphism of the elementary abelian $q$-subgroup $Q/\Phi(Q)$. By Burnside's basis theorem, $x$ continues to have order $p$ as an automorphism of $Q/\Phi(Q)$. Hence $p$ divides $|\operatorname{Aut}(Q)| = q^{c(c-1)/2}(q^c-1)(q^{c-1}-1)\cdots(q-1)$, or equivalently, $p$ divides $q^{c'} - 1$ for some $c'$ satisfying $1 \leq c' \leq c\leq b$. $\square$

Now we handle part (2).

Proposition: If $G$ has order $p^a q$ for $p,q$ distinct primes, and its Sylow $p$-subgroups are abelian, then either the Sylow $p$-subgroup is normal, or the Sylow $q$-subgroup is normal.

Proof: If the Sylow $p$-subgroup $P$ is not normal, then $N_G(P)=P$ since $P$ is a maximal subgroup. Since $P$ is abelian, $P \leq C_G(P) = N_G(P)$. Hence Burnside's normal $p$-complement theorem shows that $G$ has a normal subgroup $Q$ of index $p^a$, that is, of order $q$. Hence the Sylow $q$-subgroup of $G$, $Q$, is normal. $\square$

  • $\begingroup$ Nice examples, thanks. $\endgroup$ – Gaston Burrull Apr 27 '14 at 23:17

Given 1, how many elements of order a power of $\;p\;$ we have? At least $$\left(p^{100}-1\right)q=p^{100}q-q$$

so the remaining elements must belong to the unique Sylow $\;q$-subgroup, which is then normal

  • $\begingroup$ I did the same in the exam. So probably 1.- is false. For 2.- I dont know how to do, my attempt is proving the hipotesys for 1 to find a normal q-subgroup $\endgroup$ – Gaston Burrull Apr 18 '14 at 23:28
  • $\begingroup$ I've no idea how to prove (2) without the Feit-Thompson theorem, or if the conditions from (1) still apply then, as we have a normal Sylow $\;q$-subgroup, then (2) follows at once, but then abelianity of the Sylow $\;p$-subgroups is unnecessary... $\endgroup$ – DonAntonio Apr 18 '14 at 23:50
  • $\begingroup$ To use 2.- we must prove that there are 2 p sylow subgroups (If exists only one we are done) and for each pair of p Sylow subgroups i want to show that intersection is trivial. I dont believe that is true without abelianity. $\endgroup$ – Gaston Burrull Apr 19 '14 at 0:17

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