I was trying to solve the following problem in Isaacs' Finite Group Theory:

Let $G$ be a finite group, $\pi$ a set of primes.

a) Show that there exists a (unique) normal subgroup $N$ of $G$ such that $G/N$ is a $\pi$-group and such that it is contained in any normal subgroup $M\lhd G$ such that $G/M$ is a $\pi$-group.

b) Show that this group $N$ is generated by the set of all elements in $G$ whose order is not divisible by any prime in $\pi$.

As for the first point I think I have no problems: I define $N$ to be the intersection of all the normal subgroups which satisfy the property in a) and I prove that this intersection is characteristic in $G$ and satisfies also the same property. As for the second point, I have proved that the generated subgroup must be contained in any normal subgroup with the property in a), and so it is contained in $N$; however, I cannot prove the other inclusion. I thought that the best way was to prove that this generated subgroup is normal in $G$ and again satisfies the same property. I proved that this subgroup, which I call $S$, is characteristic in $G$, hence normal, but I don't know how to prove that the quotient group $G/S$ is a $\pi$-group. I tried to argue by contradiction: if there exists a prime $q$ not in $\pi$ which divides $|G/S|$, then by Cauchy's Thm there will be an element $xS\in G/S$ of order $q$; then by using the isomorphism $\frac{G/S}{N/S}\cong G/N$ and the fact that $G/N$ is a $\pi$-group it must be $x\in N$. But now I have no idea about how to continue in order to derive a contradiction.

Can anyone help me, please? Thank you so much for sharing your ideas!

  • $\begingroup$ I guess you should say that $G$ is solvable otherwise it is not true. Can you tell the index of question in the book? $\endgroup$
    – mesel
    Feb 24 '14 at 11:43
  • 1
    $\begingroup$ It's problem 1B8 at page 13 of the book. $\endgroup$
    – Diogenes
    Feb 24 '14 at 11:45
  • 2
    $\begingroup$ :math.umn.edu/~bahra004/fgt.pdf here is the solution of isac's book's first two cheapter.(it is solved by isacs himsel you can trust the solution :)) $\endgroup$
    – mesel
    Feb 24 '14 at 11:48
  • $\begingroup$ @mesel No, solvability is definitely not required here. $\endgroup$ Feb 24 '14 at 11:51
  • $\begingroup$ @Diogenes: for part (b) one direction is lagrange's theorem and the other is cauchy's theorem. $\endgroup$ Feb 24 '14 at 11:52

For sake of completeness, I write here what I have understood about the ideas underlying the solution of the exercise (so everybody should feel free to say if something is wrong).

a) Define $N:= \bigcap \{M \mid M\lhd G \,\,\text{and} \,\,G/M \,\,\text{is a}\,\,\pi-\text{group}\}$. Since every automorphism of $G$ preserves the order and the normality of groups, it follows that $N$ includes its image under every automorphism, and so by passing to inverse automorphisms we obtain actually equality between these two sets. So $N$ is a characteristic subgroup of $G$, in particular normal. Furthermore, by the third isomorphism theorem we have that $\frac{G/N}{M/N}\cong G/M$ for every $M\lhd G$ s.t. $G/M$ is a $\pi$-group. If by contradiction there were a prime $p\notin \pi$ which divides $|G/N|$, then by Cauchy's Theorem there would be an element $xN\in G/N$ of order $p$. But then its image $xN(M/N)\in \frac{G/N}{M/N}$ under the canonical projection would have as order a divisor of $p$, so it must be 1 because $G/M$ is a $\pi$-group. So $xN\in M/N$ and so $x\in M$. Since this holds for every $M$ which satisfies the property above, we obtain $x\in N$, and so $xN=N$ has order one, absurd.

b) Define $S:=\langle \, g\in G \mid \text{ord}(g) \,\,\text{has no prime divisor in}\,\, \pi\rangle$.

$S\subseteq N$: It suffices to show that if $g\in G$ has order not divisible by any prime in $\pi$, then it belongs to any normal subgroup $M$ such that $G/M$ is a $\pi$-group. Take $gM\in G/M$. Its order must be a divisor of the order of $g$ in $G$ and it must also be a divisor of $|G/M|$, so it must have any prime divisor simultaneously not in $\pi$ and in $\pi$. So it cannot have prime divisors, hence it must be equal to one, i.e. $gM=M$, i.e. $g\in M$.

$N\subseteq S$: By the property of point a) it suffices to show that $S\lhd G$ and $G/S$ is a $\pi$-group. By the same argument of above, the automorphisms of $G$ preserve orders, so it follows that the subgroup $S$ is characteristic in $G$, hence normal. It suffices to show that $G/S$ is a $\pi$-group. Take a prime divisor $p$ of $|G/S|$. By Cauchy's theorem there exists an element $gS\in G/S$ of order $p$, so that $g^p\in S$ is the least positive power of $g$ which belongs to $S$. Write now $\text{ord}(g)=p^kn$, for some natural numbers $k$ and $n$ with $\text{gcd}(p,n)=1$. By this extremely powerful weapon called Bézout's identity, there exist integers $a,b\in \mathbb{Z}$ such that $ap+bn=1$, and so we have $g^{ap+bn}=g\notin S$, and so it must be also $g^n\notin S$ (because we already know that $g^{ap}\in S$). But now we have $g^n\notin S$, so by definition of $S$ we have that $\text{ord}(g^n)=p^k$ must have at least one prime divisor in $\pi$, and this prime divisor can only be $p$ itself. Hence $p\in \pi$, and so $G/S$ is a $\pi$-group.

  • $\begingroup$ Looks good. :-) $\endgroup$ Feb 24 '14 at 15:59
  • $\begingroup$ @Diogenes: A beautiful argument! Thank you for taking the time to answer your own question. $\endgroup$
    – Moritz
    Aug 25 '18 at 12:44

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