# sylow basis of finite solvable groups

Let ‎‎$G‎‎$‎ be a‎ ‎finite ‎solvable non-$p$-group ‎and ‎‎$‎‎A$ ‎be a ‎‎‎‎maximal ‎subgroup ‎of ‎‎$G‎‎$‎. ‎ Therefore ‎$A‎‎$ ‎is ‎of ‎primary ‎index ‎‎$p^{n}‎$‎‎, that is ‎$|G : A|=p^{n}‎$ ‎‎where ‎$‎‎p$ ‎is ‎prime ‎and‎ ‎‎$n\geq 1‎‎$‎‎. ‎ ‎If ‎$‎‎|G|=p_1^{n_1}p_2^{n_2}\cdots p_l^{n_l}p^{m+n}$‎; ‎$‎‎n_i\geq 1, m\geq 0$ ‎‎ and ‎‎$P_i‎‎$ ‎is a sylow ‎$‎‎p_{i}$-‎subgroup‎‎ ‎and ‎$‎‎P$ is a sylow ‎$‎‎p$-‎subgroup of ‎$‎‎G$‎‎‎‎, ‎ c‎an we obtain a sylow ‎basis ‎$\{P_1, P_2, \cdots,P_l, P\} ; ‎‎ ‎l\geq 1‎$ ‎of ‎‎$G‎‎$ ‎such ‎‎ that ‎‎$P_i‎\subseteq‎‎ ‎A‎$; ‎$‎‎1\leq i\leq l$‎‎? ‎ Any answer will be appreciated

This is a special case of Hall (1937), and is mostly easily understood after you've seen Hall subgroups (subgroups whose order is coprime to their index). You can see the ideas begin in Hall (1928), but you can also see them put together in a nice way in Carter (1959). In particular Carter describes how to use the maximality condition in a non-trivial way to understand subgroups of a finite soluble group. The proof below does not use maximality and holds for any subgroup $A$ of a finite soluble group $G$ (except for the part where $\tilde P_i = P_i$).

## Main construction

Set $p_0 = p$ and let ‎$\{\tilde P_1, \tilde P_2, \cdots, \tilde P_l, \tilde P = \tilde P_0\}$ be a Sylow basis for $A$. Let $\tilde Q_i$ be the product of the $\tilde P_j$ for $j \neq i$. Since we have a Sylow basis, each $\tilde Q_i$ is a subgroup of order coprime to $p_i$ and of index (relative to $A$) a power of $p_i$ (called a Hall $p_i'$-subgroup, or a Sylow $p$-complement; the collection of $\{\tilde Q_i\}$ is called a Sylow complement system for $A$).

Let $Q_i$ be a Hall $p_i'$-subgroup of $G$ containing $\tilde Q_i$ (let me know if you need to see a proof of the existence of $Q_i$; this is in Hall's 1928 paper). Define $P_\pi = \bigcap_{p_j \notin \pi} Q_j$. Then I claim $\{ P_i \} = \{ P_{\{p_i\}} \}$ is a Sylow basis of $G$, and that it extends the Sylow basis of $A$.

## Verification

• $P_\pi$ is a $\pi$-subgroup.

Proof: If $p_j \notin \pi$, then $P_\pi \leq Q_j$, and $Q_j$ is a $p_j'$-subgroup, so $P_\pi$ does not have order divisible by $p_j$.

• $P_\pi$ is a Hall $\pi$-subgroup of $G$.

Proof: This is true by definition if $\pi$ is missing only zero or one primes dividing $|G|$, since then $P_\pi = G$ or $P_\pi = Q_j$. Suppose for induction's sake that $P_\pi$ is a Hall $\pi$-subgroup of $G$ for $\pi$ missing less than $k$ primes dividing the order of $G$, and let $\pi = \rho \setminus \{ p_i \}$ for $\rho \neq \pi$ be a set of primes missing exactly $k$ prime divisors of $|G|$. Then by definition $P_\pi = P_\rho \cap Q_i$. Since $P_\rho$ is a Hall $\rho$-subgroup by induction, and $Q_i$ is a $p_i'$-subgroup, we have $[G:P_\rho]$ is a $\rho'$-number and $[G:Q_i]$ is a $p_i$-number, so their indices are relatively prime and hence $[G:P_\pi] = [G:P_\rho] [G:Q_i]$ so that $P_\pi$ is a Hall $\pi$-subgroup of $G$, and the induction proceeds.

• $P_\pi \cap P_\rho = P_{\pi \cap \rho}$

Proof: This is just the definition.

• $P_\pi P_\rho = P_\rho P_\pi = P_{\rho \cup \pi}$

Proof: $P_\pi$ and $P_\rho$ are by definition subgroups of $P_{\rho \cup \pi}$, so that $P_\pi P_\rho$ is a subset of $P_{\rho \cup \pi}$. However $|P_\rho P_\pi| = |P_\rho| |P_\pi| / |P_\rho \cap P_\pi|$ which is exactly the size of $|P_{\rho \cup \pi}|$, so we have $P_\pi P_\rho = P_{\rho \cup \pi}$ is a group.

• $A \cap P_i = \tilde P_i$. (That is the Sylow basis of $G$ that we constructed “reduces” into $A$)

Proof: $A \cap P_i = A \cap \bigcap_{j \neq i} Q_j = \bigcap{j \neq i} (A \cap Q_j) = \bigcap_{j \neq i} \tilde Q_j = \tilde P_j$. The only tricky equality is the last: by definition $\tilde P_i \leq \tilde Q_j$ so we get the $\supseteq$. The other inclusion is just an order consideration using the previous claims for $\tilde P_\pi = \bigcap_{p_j \notin \pi} \tilde Q_j$.

• $P_i = \tilde P_i$ if $i \neq 0$.

Proof: This is just an order consideration. $P_i$ is a Sylow $p_i$-subgroup of $G$ containing $\tilde P_i$, a Sylow $p_i$-subgroup of $A$, but the same power of $p_i$ divides both $|G|$ and $|A|$.

## Bibliography

• Hall, P. “A note on soluble groups.” Journal of the London Mathematical Society 3, (1928) 98-105. JFM 54.0145.01 DOI:10.1112/jlms/s1-3.2.98

• Hall, P. “On the Sylow systems of a soluble group.” Proceedings of the London Mathematical Society, (2nd series) 43, (1937) 316-323. JFM 63.0069.04 DOI:10.1112/plms/s2-43.4.316

• Carter, R. W. “On a class of finite soluble groups.” Proceedings of the London Mathematical Society, (3rd series) 9 (1959) 623–640. MR114859 DOI:10.1112/plms/s3-9.4.623

• Short version: Sylow bases are in bijection with Sylow complement bases, and Sylow complement bases are extendable. – Jack Schmidt May 15 '13 at 20:54
• Dear professor Schmidt many thanks for your exact and nice answer. It is comprehensive as usual! – sebastian May 16 '13 at 5:26