1.11. Sylow Theorems.
Lagrange's Theorem states that \(H \le G\), then \(|H|\) divides \(|G|\). However, you may wonder if there is some kind of converse. If \(k\) divides \(|G|\), is there a subgroup of order \(k\)?
By Cauchy's Theorem, we know that if \(p\) is a prime which divides then there exists an element of order \(p\). Can we generalize this result further (for example, state how many such elements satisfy this)?
The answer to both questions is yes and is achieved through Sylow's Theorem. It's a foundational theorem in finite group theory, as it strengthens our two most power theorems for finite groups: Lagrange's Theorem and Cauchy's Theorem.
\(H\) is a \(p\)-subgroup of a group \(G\) if \(H\) is a subgroup of \(G\) and \(|H| = p^n\) for some \(n \ge 1\).
Let \(G\) be a group and let \(p\) be a prime such that \(p\mid |G|\). Suppose \(p^k\) is the largest power such that \(p^k \mid |G|\). That is, \(|G| = p^km\) for some integer \(m \in G\), \(\mbox{gcd}(p, m) = 1\). Then any subgroup \(H\) of \(G\) with \(|H| = p^k\) is called a Sylow \(p\)-subgroup. \ \ An equivalent definition is the following: \(H\) is a Sylow-\(p\) subgroup if \(H\) is a \(p\)-subgroup where \(|H| = p^k\).
\textcolor{purple}{A Sylow \(p\)-subgroup is nothing more than a subgroup \(H\) where \(|H| = p^k\) and \(|G| = p^km\) where \(\mbox{gcd(p, m)} = 1\).}
Let \(G\) be a group and \(P\) and \(Q\) be subgroups of \(G\). If there exists an element \(g \in G\) such that
then \(P\) is conjugate to \(Q\). Recall that if \(H \normal G\), then for any \(g \in G\) we see that \(gHg^{-1} = H\). Thus \(H\) is conjugate to itself. Also note that if \(P\) is a subgroup then so is \(gPg^{-1}\).
[ (Sylow Theorem)] Let \(G\) be a finite group and \(p\) a prime such that \(p \mid |G|\). Suppose further that \(|G| = p^km\) where \(\mbox{gcd}(p, m) = 1\). Then
- 1. There exists a Sylow \(p\)-subgroup (equivalently, there exists a subgroup \(H\) of \(G\) where \(H = p^k\)) and every \(p\)-subgroup of \(G\) is contained in some Sylow \(p\)-subgroup
- 2. All Sylow \(p\)-subgroups are conjugate to each other, and the conjugate of any Sylow \(p\)-subgroup is also a Sylow \(p\)-subgroup
- 3. If \(n_p\) is the number of Sylow \(p\)-subgroups, then
\vspace{-.4cm}
\textcolor{NavyBlue}{We can prove the first part by letting \(G\) act on a special set \(\Omega\). It will turn out that the stabilizer of our action will be the desired Sylow \(p\)-subgroup.}
- 1. Define
and let \(G\) act on \(\Omega\) from the left. Observe that for \(X \in \Omega\), \(g * X = \{gx \mid \text{ for all } x \in X\} = gX.\) Since \(|gX| = |X| = p^k\), we see that \(gX \in \Omega\). Associativity and identity applications are trivial, so we get that this is a group action.
\textcolor{NavyBlue}{Now that we have shown that this is a group action, we will consider the orbits of the group aciton.}
Since \(|G| = p^km\), there are \(\displaystyle \binom{p^km}{p^k}\) many ways for us to choose a subset \(X\) of \(G\) with size \(p^k\). Hence \(|\Omega| = \displaystyle \binom{p^km}{p^k}\). Note that since this is a group action, the orbits form a partition of \(X\). Now from number theory, we know that
Since the orbits must partition \(\Omega\), the above result tells us that we cannot partition \(G\) with sets which are divisible by \(p\). In other words, there must exist some orbit \(\mathcal{O}\) such that \(|\mathcal{O}|\) is not divisible by \(p\).
\textcolor{NavyBlue}{Now that we know that there exists an orbit not divisible by \(p\), we will anaylze the corresponding stabilizer of this orbit. This stabilizer will turn out to be our Sylow \(p\)-subgroup. }
Let \(H\) be the orbit corresponding to \(\mathcal{O}\). Then by the Orbit-Stabilizer Theorem
By the last equation, we see that \(p^k\) must divide both sides. However, \(|\mathcal{O}|\) is not divisible by \(p\). Hence \(|H|\) must be divisible by \(p^k\).
However, by Lagrange's Theorem, \(|H|\) divides \(|G| = p^km\). Therefore \(|H| = m\) or \(|H| \in \{1, p, p^2, \dots, p^k\}\). In either case \(|H| \le p^k\) (since \(m \le p\)). But we just showed that \(p^k\) divides \(|H|\), which proves that \(|H| = p^k\).
Since \(H\) is a stabilizer, \(H \le G\), so we have effectively proved the existence of a subgroup of order \(p^k\); or, in other words, a Sylow \(p\)-subgroup. * 2. Suppose \(H\) and \(K\) are Sylow \(p\)-subgroups of \(G\). Then observe that * 3.
The consequences of this theorem are immediate.
Let \(G\) be a finite group and suppose \(|G| = p^km\) for some prime \(p\) where \(\mbox{gcd}(p, m) = 1\). Then \(G\) has a normal subgroup of order \(p^k\) if and only if \(n_p = 1\).
(\(\implies\)) Suppose \(G\) has a normal subgroup \(H\) of order \(p^k\). By Sylow's Theorem, we know that all other Sylow \(p\)-subgroups are conjugate to \(H\). Thus let \(g \in G\) and observe that
since \(H\) is normal. Therefore, there are no other Sylow \(p\)-subgroups so \(n_p = 1.\)
(\(\impliedby\)) Now suppose that \(n_p = 1\). Let \(H\) be a sole Sylow \(p\)-subgroup of \(G\). Since it is the only Sylow \(p\)-subgroup, we see that
for all \(g \in G\). However this exactly the definition for \(H\) to be a normal subgroup of \(G\). This proves the result.
Once you use Sylow's Theorem and study finite groups more, you'll realize that some groups aren't that complicated. For example, consider any subgroup of order 4. This can be any wild group you want, but at the end of the day, it turns out one of the following options is true:
The process leading to such a conclusion is known as classifying groups up to an isomorphism. That is, you start with a group with a fixed order, and then determine much simpler groups that your group could be isomorphic to. In our example, we say that any group of order 4 can only be two things up to an isomorphism.
The cool thing about Sylow's Theorem is that it is so strong that it allows us to classify groups up to an isomorphism.
In general, when classifying groups up to an isomorphism, it is convenient to do in terms of integer groups \(\mathbb{Z}\) or modulo integer groups, as we saw above. This isn't always possible, but when it is, the following theorem comes in handy.
Let \(m, n\) be positive integers. Then
if and only if \(m\) and \(n\) are coprime.
\noindent Example. \ \textcolor{NavyBlue}{Suppose we want to classify all groups of order 1225 up to an isomorphism.} \ Let \(G\) be a group such that \(|G| = 1225 = 5^27^2\). Then observe \(\mbox{gcd}(5, 7^2) = 1\). By Sylow's theorem, we know that if \(n_5\) is the number of Sylow \(5\)-subgroups of \(G\), then
Observe that \(n_5\) can only equal 1. Since \(n_5 = 1\), we know by Propsition \ref{sylow_normal} that for the unique Sylow 5-subgroup \(H\) that \(H \unlhd G\). Also note that \(|H| = 5^2\). \ \ Now observe that \(\mbox{gcd}(7, 5^2) = 1\). By Sylow's Theorem, we know that if \(n_7\) is the number of Sylow 7-subgroups of \(G\) that
Note that \(n_7\) must also equal 1. Thus again for the unique Sylow 7-subgroup \(K\), we must have that \(K \unlhd G\) and \(|K| = 7^2\). Now we can observe that (1) \(\mbox{gcd}(|H|, |K|) = 1\) and (2) \(|G| = |H||K|\) so that
by Theorem 1.\ref{product_theorem}. Now observe that since \(|H| = 5^2\), \(H \cong \mathbb{Z}/25\mathbb{Z}\) and \(H \cong \mathbb{Z}/5\mathbb{Z} \times \mathbb{Z}/5\mathbb{Z}\). Since \(K = 7^2\), \(K \cong \mathbb{Z}/49\mathbb{Z}\) and \(H \cong (\mathbb{Z}/7\mathbb{Z} \times \mathbb{Z}/7\mathbb{Z})\). Therefore, we see that the groups of order 1225 are, up to isomorphism,
- (1) \((\mathbb{Z}/25\mathbb{Z}) \times (\mathbb{Z}/49\mathbb{Z})\)
- (2) \((\mathbb{Z}/25\mathbb{Z}) \times (\mathbb{Z}/7\mathbb{Z} \times \mathbb{Z}/7\mathbb{Z})\)
- (3) \((\mathbb{Z}/5\mathbb{Z} \times \mathbb{Z}/5\mathbb{Z}) \times \mathbb{Z}/49\mathbb{Z}\)
- (4) \((\mathbb{Z}/5\mathbb{Z} \times \mathbb{Z}/5\mathbb{Z}) \times (\mathbb{Z}/7\mathbb{Z} \times \mathbb{Z}/7\mathbb{Z})\).
\textcolor{purple}{We suspect that these are all the groups of order 1225 up to an isomorphism. However, we double check that none of these groups are actually equivalent to each other, i.e., that we have no redundancies.}
Observe that (1) is not isomorphic to any of the the other groups, since \((1, 1) \in (\mathbb{Z}/25\mathbb{Z}) \times (\mathbb{Z}/49\mathbb{Z})\), has order 1225 but none of the other groups have an element of order 1225. \ \ In addition, (3) is not isomorphic to (2) or (3) since \((0, 1) \in (\mathbb{Z}/5\mathbb{Z} \times \mathbb{Z}/5\mathbb{Z}) \times \mathbb{Z}/49\mathbb{Z}\) and has order 49 but no element of either (2) or (3) has an element of either 49. \ \ Finally, we see that (2) is not isomorphic to (4) because \((1, 0) \in (\mathbb{Z}/25\mathbb{Z}) \times (\mathbb{Z}/7\mathbb{Z} \times \mathbb{Z}/7\mathbb{Z})\) is an element of order 25 but there is no element of order 25 in (4). Thus we see that (1) these subgroups are isomorphic to \(G\) and (2) none of them are isomorphic to each other. Therefore, this an exhaustive list of all the groups of order 1225 up to isomorphism.
Here's another example in which Sylow's Theorem helps us classify a specific type of group.
Let \(p,q\) be primes with \(p<q\) and suppose \(p\) does not divide \(q-1\). If \(G\) is a group such that \(|G| = pq\), then \(G \cong \mathbb{Z}/pq\mathbb{Z}\).
Let \(G\) be a group and \(|G| = pq\). Since \(\mbox{gcd}(p, q) = 1,\) by the Sylow Theorem, there exists a Sylow \(p\)-subgroup and Sylow \(q\)-subgroup of \(G\). \ Now let \(n_p\) and \(n_q\) be the number of Sylow \(p\) and \(q\)-subgroups, respectively. Then observe that
so that \(n_p = 1\) and
Now observe that \(n_p = 1\) or \(q\). However, since \(p\) does not divide \(q - 1\), we know that
Thus \(n_p = 1\). Again, either \(n_p = 1\) or \(p\) but \(p < q\) so
unless \(n_q = 1\). Thus there is one and only one Sylow \(p\)-subgroup and Sylow \(q\)-subgroup, which we can call \(H\) and \(K\) respectively. By proposition \ref{sylow_normal},
Note that (1) \(\mbox{gcd}(|H|, |K|) = \mbox{gcd}(p, q) = 1\) and (2) \(|G| = |H||K| = pq\). Thus \(G \cong H \times K\) by Theorem 1.\ref{product_theorem}. Now observe that \(H\) and \(K\) are of prime order, so that \(H \cong \mathbb{Z}/p\mathbb{Z}\) and \(K \cong \mathbb{Z}/q\mathbb{Z}\). We then see that
From theorem ???, we know that if \(m, n\) are positive integers and \(\mbox{gcd}(m, n) = 1\), then
Obviously, \(\mbox{gcd}(p, q) = 1\), so that
Now isomorphic relations are transitive, so we can finally state that
as desired.