Lagrange's theorem says which subgroup orders are possible but promises none, and its converse fails. For prime powers, though, there is a strong partial converse. The Sylow theorems, proved with group actions, are the main tool for understanding groups of a given order.
Write $|G| = p^{a}m$ with $p$ prime and $p \nmid m$. A subgroup of order $p^{a}$, the largest power of $p$ dividing $|G|$, is a Sylow $p$-subgroup.
- Sylow I (existence). $G$ has a Sylow $p$-subgroup. (Hence, by Cauchy's theorem, $G$ has an element of order $p$ for every prime $p$ dividing $|G|$.)
- Sylow II (conjugacy). Any two Sylow $p$-subgroups are conjugate: $Q = gPg^{-1}$ for some $g$.
- Sylow III (counting). The number $n_{p}$ of Sylow $p$-subgroups satisfies $n_{p} \mid m$ and $n_{p} \equiv 1 \pmod{p}$.
Why it works
Worked example: Order 20
- $20 = 5 \cdot 4$, so $n_{5}$ divides $4$ and $n_{5} \equiv 1 \pmod 5$.
- Divisors of $4$: $1, 2, 4$. Only $1$ is $\equiv 1 \pmod 5$.
- So $n_{5} = 1$, and the Sylow 5-subgroup is normal.
Answer: Yes.
Worked example: Order 15 is cyclic
- $n_{3} \mid 5$ and $n_{3} \equiv 1 \pmod 3$: $n_{3} = 1$. $n_{5} \mid 3$ and $n_{5} \equiv 1 \pmod 5$: $n_{5} = 1$.
- So there is one subgroup $P$ of order 3 and one, $Q$, of order 5; they hold $2 + 4 = 6$ non-identity elements.
- The other $15 - 1 - 6 = 8$ elements have order dividing $15$ but not $1$, $3$ or $5$: order $15$. So the group is cyclic.
Answer: $G \cong \mathbb{Z}_{15}$.
Worked example: Order 12
- $12 = 3 \cdot 4$: $n_{3} \mid 4$ and $n_{3} \equiv 1 \pmod 3$.
- Candidates $1, 2, 4$: $n_{3} \in \{1, 4\}$. Both happen: $\mathbb{Z}_{12}$ has $n_{3} = 1$, and $A_{4}$ has $n_{3} = 4$ (the subgroups generated by its eight 3-cycles, two per subgroup).
Answer: $n_{3} \in \{1, 4\}$.
Counting elements is a frequent finishing move: Sylow subgroups of prime order $p$ meet only in $e$, so $n_{p}$ of them hold $n_{p}(p - 1)$ elements of order $p$. If these counts add up to more than $|G|$, some $n_{p}$ must be $1$.
Common mistake
Key idea