2.5 Sylow Theorems
Sylow Theorems provide a partial converse to Lagrange’s Theorem.
Definition 2.5.1. Let \(G\) be a finite group and \(p\) be a prime. Any subgroup of \(G\) whose order is the
highest power of \(p\) dividing \(\begin {vmatrix} G\\ \end {vmatrix}\) is called a \(p-\)sylow subgroup of \(G\).
A \(p-\)sylow subgroup for some \(p\) is called a Sylow subgroup.
- 1.
- Consider a group of order 100
\(100 = 2^2\cdot 5^2\), the group has a \(2-\)sylow subgroup (of order 4) and \(5-\)sylow subgroup (of order 25). - 2.
- For a group of order 12,
\(12 = 2^3\cdot 3\), \(2-\)sylow subgroups of order 4 and \(3-\)sylow subgroups of order 3. - 3.
- Consider \(\mathbb {Z}_{12}.\)
\(\mathbb {Z}_{12} = \{0\, , \, 1\, , \, 2\,, \cdots \cdots \cdots ,11\}\,\) is \(\, \{0\, , \, 3\, , \, 6\, , \,9\} = \langle 3\rangle \)
The \(3-\)sylow subgroups of order 3 is \(\, \{0\,,\, 4\, , \, 8\} = \langle 4\rangle \). - 4.
- Consider \(A_4\).
\(\begin {vmatrix} A_4\\ \end {vmatrix} = 12\). We know that \(A_4\) has a subgroup of order 4.
\(P-\)Sylow Subgroups of a Group \(G\)
Sylow\(-P\)
\(A_4\) has order 12.
\(12 = 2^2\cdot 3\)
There is one group of order 4, so the only \(2-\)sylow subgroups of \(A_4\) is
\[\, \{(1)\, , \, (12)(34)\, , \, (13)(24)\, , \, (14)(23)\} = \langle (12)(34)\, , \, (14)(23)\rangle \]
There are four \(3-\)sylow subgroups; \begin {align*} \{(1)\, , \, (123)\, , \, (132)\} & = \langle (123)\rangle \\ \{(1)\, , \, (124)\, , \, (142)\} & = \langle (124)\rangle \\ \{(1)\, , \, (134)\, , \, (143)\} & = \langle (134)\rangle \\ \{(1)\, , \, (234)\, , \, (243)\} & = \langle (234)\rangle \\ \end {align*}
We first start with a weak form of the Sylow theorems.
Proposition 2.5.3. If \(G\) is a finite \(A\) belian group and \(p\) is a prime dividing the order of \(G\), then \(G\)
has an element of order \(p\).
First Sylow Theorem
Let \(G\) be a finite group, \(p\) a prime and \(p^r\) the highest power of \(p\) dividing the order of \(G\). Then there is a
subgroup of order \(p^r\).
Second Sylow Theorem
Let \(H\) be a subgroup of a finite group \(G\), and let \(p\) be a Sylow \(p\)-subgroup of \(G\). If \(H\) is a \(p-\)group, then \(H\) is
contained in a \(G-\)conjugate of \(p\).
Third Sylow Theorem
- (i)
- Any two sylow \(p-\)subgroups of a finite group \(G\) are conjugates.
- (ii)
- The number \(S_p\) of distinct sylow \(p-\)subgroups of \(G\) is congruent to 1 modulo \(p\).
- (iii)
- \(S_p\big / \begin {vmatrix} G\\ \end {vmatrix}\)
Proof.
- (i)
- Let \(p\) and \(p'\) be two sylow \(p-\)subgroups. Then by the \(2^{\text {nd}}\) sylow theorem, \(p'\) as a \(p-\)group is contained in some \(G-\)conjugate \(R\) of \(p\). But \(\, \begin {vmatrix} p\\ \end {vmatrix} = \begin {vmatrix} p'\\ \end {vmatrix} \implies p = p'\).
- 1.
- Let \(p\) be any Sylow \(p-\)subgroup. Since any other sylow \(p-\)subgroup is conjugate to \(p\) and conjugate
of a sylow \(p-\)subgroup is a sylow \(p-\)subgroup, we conclude that
\(S_p = \big [G:\, N_G(p)\big ]\). But on putting \(p=H\) in the following: Recall \[\, \big [G:\, N_G(p)\big ] = \sum \big [H:\, H\cap R\big ]\] Thus we have \(S_p = \sum \limits _{\textit {R}\in \textbf {R}}\big [R:\, P\cap R\big ]\).
Now, for exactly one \(R\in \textbf {R}\, , \, R=p\) for every \(p-\)conjugate of \(p\) is itself and so \(p\) is the only possible representative of its equivalence class. In all other cases \(p\cap R\neq p\). Therefore \([p:\, p\cap R]\) is a power of \(p\) for all \(R\in \mathbb {R}\) except one, and for this one, \([p:\, p\cap R] = 1\). Hence \(S_p = 1 + R_p = 1\) mod\(_p\).
Summary of Sylow Theorems
- •
- If \(\begin {vmatrix} G\\ \end {vmatrix} = p^rm\), where \(p\) does not divide \(m\), then \(G\) has a subgroup of order \(p^r\).
- •
- The number \(S_p\) of sylow \(p-\)subgroup of \(G\) satisfy \(S_p = 1\) mod\(_p\).
- •
- \(S_p\big /\begin {vmatrix} G\\ \end {vmatrix}\,\) further \(\, S_p\big /m\).
- •
- Any two sylow \(p-\)subgroups are conjugate.
Consequences of the Sylow Theorems and Applications
- •
- If a Sylow \(p-\)subgroups is unique, then it is a normal subgroup [check example on how to
count \(S_p\)].
Example 2.5.4. Consider \(G\) such that \(\begin {vmatrix} G\\ \end {vmatrix} = 50\)
\(\begin {vmatrix} G\\ \end {vmatrix} = 50 = 5^2\cdot 2\)
\(S_2 = 1\) or \(5\) or \(25\). since \(S_2\) divides \(\begin {vmatrix} 5^2\\ \end {vmatrix} = 25\) and \(\equiv 1\) mod\(_2\)
\(S_5 = 1\) since \(S_5\) divide \(\begin {vmatrix} 2\\ \end {vmatrix}\) and \(\equiv 1\) mod\(_5\)
- •
- If \(G\) has order \(2p\), where \(p>2\). Then \(G\) is isomorphic to \(\mathbb {Z}_{2p}\) or \(D_p\)
- •
- If \(\begin {vmatrix} G\\ \end {vmatrix} = pq,\, \, p\, , \, q\) primes, \(p<q\) and that \(p\) does not divide \(q-1\), then \(G\) is Cyclic. In particular \(G\) is isomorphic to \(\mathbb {Z}_{pq}\).
- •
- Classification of groups. Simple/not simple.
- •
- Small \(s = \) number of Sylow groups.
- •
- Primary test
One such theorem used to test if a given number is prime is WILSON’S THEOREM.
Wilson’s Theorem
Given \(\, p\in \mathbb {N}\), then \(p\) is prime if and only if \((p - 1)! = -1\) mod\(_p\).
Proof. Consider the group Sym\(_p\) on \(p\) elements. \(\begin {vmatrix} S_p\\ \end {vmatrix} = p! = p(p-1)!\)
\(S_p = (p-1)! = -1\) mod\(_p\)
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