I went to search info about it in books (as Isaac), this site, wikipedia. But no one says nothing about my (maybe silly) doubt.
The theorem states that, given a finite and solvable group $G$, and an arbirary set of primes $\pi$, there will always exists a subgroup of $G$, call it $H$, such that:
- $|H|$ is a $\pi$-number, i.e. every prime divisor of $|H|$ is in $\pi$.
- $|G:H|$ is a $\pi '$-number, i.e. every prime divisor of $|G:H|$ is not in $\pi$.
But I can't understand how can this is possible if, for example, the smallest prime in $\pi$ is larger then the greatest prime divisor of $|G|$.
Take for example $\pi=\{17\}$ and $G=\langle a\rangle$ the cyclic group of order $13$. $G$ is abelian, hence solvable. Its subgroups are clearly only the trivial ones, hence it seems the theorem fails.
Even the proof: it's done by induction. And it works in the inductive step... but not in the base case, when $|G|=1$ because $H\leq G\Rightarrow H=1$ and $1=|H|=|G:H|$ so both 1) and 2) fails, unless we assume that by agreement $1$ is both a $\pi$-number and a $\pi '$-number.
Could someone explain this? I would be really grateful.
(I hope to have wrote nothing blasphemous!)
1 is a $\pi$-number for all sets of primes $\pi$.
If $\pi$ is a set of primes and $n$ is an integer (usually positive), then $n$ is a $\pi$-number if and only if every prime divisor of $n$ is contained in $\pi$. One can define $\pi(n)$ to be the set of prime divisors of $n$ (though I find this a disconcerting use of $\pi$ to mean both a set and an unrelated function), and then $n$ is a $\pi$-number if and only if $\pi(n) \subseteq \pi$. For a finite group $G$, we say that $G$ is a $\pi$-group if and only if its order is a $\pi$-number. One can define $\pi(G) = \pi(|G|)$ (again unrelated to the set), and then a finite group is a $\pi$-group if and only if $\pi(G) \subseteq \pi$. A finite subgroup $H \leq G$ is called a $\pi$-subgroup if $H$ is a $\pi$-group. A finite order element $g$ of a group is a $\pi$-element if its order is a $\pi$-number (I believe elements of infinite order are not $\pi$-elements for any set of primes $\pi$, though I am less certain if $\pi$ is the set of all primes). A group is a $\pi$-group if and only if all of its elements are $\pi$-elements.
As an example, 1, 2, 3, 4, 6, 8, 9, 12 are the first eight positive $\{2,3\}$-numbers even though $\pi(1)=\{\} \subsetneq \{2,3\}$ and $\pi(2)=\{2\} \subsetneq \{2,3\}$.
The complement of $\pi$ is $\pi' = \{ p : p \text{ prime}, p \notin \pi \}$.
A Hall $\pi$-subgroup $H \leq G$ is a subgroup where $H$ is a $\pi$-group and $[G:H]$ is a $\pi'$-number. These are most useful in solvable groups where they (a) exist, (b) contain all $\pi$-subgroups (that is, they are the maximal $\pi$-subgroups), and (c) are all conjugate.
Some reasons that we use these definitions of $\pi$-group: $\pi$-groups are closed under subgroups and quotient groups. It would be incredibly inconvenient if $1$ was not a $\pi$-number. Consider $O^{\pi}(G)$, the smallest normal subgroup $N$ such that $G/N$ is a $\pi$-group. The way the definitions are setup, $O^{\pi}(G)$ is a characteristic subgroup. If one chose to make $1$ not a $\pi$-number, then $O^{\{3\}}(S_3)$ would not exist. Similarly, $O_p(G)$ is the name of the subgroup defined by these two equivalent definitions: (a) the intersection of the Sylow $p$-subgroups, and (b) the largest normal $p$-subgroup. If $1$ were not a $\{p\}$-number, then $O_2(S_3)$ would not exist by definition (b).