Hall subgroups exist in finite solvable group

Statement

Suppose $G$ is a finite solvable group, and $\pi$ is a prime set. Then, there exists a $\pi$-Hall subgroup (?) of $G$: a subgroup whose order and index are relatively prime, and with the property that the set of prime divisors of its order is within $\pi$.

Related facts

• Hall's theorem is a converse of sorts.
• Sylow's theorem: Sylow subgroups exist, Sylow implies order-conjugate, Sylow implies order-dominating, congruence condition on Sylow numbers
• ECD condition for pi-subgroups in solvable groups: Hall subgroups exist in finite solvable, Hall implies order-conjugate in finite solvable, Hall implies order-dominating in finite solvable, congruence condition on factorization of Hall numbers
• Pi-Hall subgroups exist in pi-separable

Facts used

1. Solvability is subgroup-closed
2. Solvability is quotient-closed
3. Minimal normal implies elementary Abelian in finite solvable
4. Sylow subgroups exist
5. Normality is quotient-transitive
6. Frattini's argument
7. Product formula

Proof

Given: A finite solvable group $G$, a prime set $\pi$.

To prove: $G$ has a $\pi$-Hall subgroup.

Proof: We prove the claim by induction on the order of $G$. The base case of the trivial group is clear.

Suppose the result is true for all finite solvable groups of order strictly less than the order of $G$. Then, by facts (1) and (2), the result is true for all proper subgroups and all proper quotients of $G$.

Let $N$ be a minimal normal subgroup of $G$. By fact (3), $N$ is elementary Abelian, and in particular, is a $p$-group for some prime divisor of $G$. Observe that:

1. Case $p\in \pi$: By applying induction to $G/N$, we conclude that $G/N$ has a $\pi$-Hall subgroup. Taking the inverse image of this in $G$, we get a $\pi$-Hall subgroup of $G$.
2. Case $p\notin \pi$: By applying induction to $G/N$, we conclude that $G/N$ has a $\pi$-Hall subgroup. The inverse image of this gives a subgroup, say $K$, of $G$. If $G/N$ is not itself a $\pi$-group, then $K/N$ is proper in $G/N$, and $K$ is a proper subgroup of $G$, then $K$ has a $\pi$-Hall subgroup $H$. Note that $[G:K]=[G/N:K/N]$ is relatively prime to $\pi$, and $[K:H]$ is relatively prime to $\pi$, so $[G:H]$ is relatively prime to $\pi$. Thus, $H$ is a Hall $\pi$-subgroup of $G$.

Thus, the only case of concern is where $p\notin \pi$ but $G/N$ is a $\pi$-group. Note that in this case, $N$ must be a $p$-Sylow subgroup (since $G/N$ has order relatively prime to $p$).

Let $M$ be a subgroup of $G$ such that $M/N$ is a minimal normal subgroup of $G/N$. Then, since the order of $G/N$ is not a multiple of $p$, and $G/N$ is solvable, fact (3) yields that $M/N$ is an elementary Abelian $q$-group for some prime $q\neq p$. By fact (5), $M$ is itself a normal subgroup of $G$.

Let $Q$ be a $q$-Sylow subgroup of $M$ (using fact (4)).

Now, if $Q$ is normal in $G$, then argue as before replacing $N$ by $Q$. Note that by our assumption, $q\in \pi$, so we land up in case (1).

Let's now consider the case that $Q$ is not normal in $G$. By Frattini's argument (fact (6)), $MN_{G}(Q)=G$. The product formula yields:

${\frac {|G|}{|N_{G}(Q)|}}={\frac {|M|}{|M\cap N_{G}(Q)|}}$

$Q\leq M\cap N_{G}(Q)$, so $|M|/(|M\cap N_{G}(Q)|)$ divides $|M|/|Q|$, and is hence a power of $p$. Thus, $|G|/|N_{G}(Q)|$ is a power of $p$. Thus, $N_{G}(Q)$ is a proper subgroup of $G$ whose index is a power of $p$. By induction, $N_{G}(Q)$ has a $\pi$-Hall subgroup, and this must also be a $\pi$-Hall subgroup for $G$.

References

Textbook references

• Abstract Algebra by David S. Dummit and Richard M. Foote, 10-digit ISBN 0471433349, 13-digit ISBN 978-0471433347, ^{More info}, Page 200, Exercise 33, Section 6.1 ($p$-groups, nilpotent groups and solvable groups)