Theorem 1.22.1. Cauchy’s theorem.
Let \(G\) be a finite group, and let \(p\) be a prime divisor of \(\abs{G}\text{.}\) There is an element of \(G\) of order \(p\text{.}\)
Section 1.22 Sylow theorems: applications
Much of the rest of this course will be devoted to classifying finite groups up to isomorphism, and the Sylow theorems will be an important tool in this regard. We illustrate this with some first results here.
Proof.
Let \(H\in Syl_p(G)\) and take any nontrivial element \(g\in H\text{.}\) Since \(\abs{g}\) is cyclic of cardinality \(p^i\) for some \(1\leq i\text{,}\) there is an element \(\abs{g}\) of order \(p\text{.}\)
Theorem 1.22.2. Groups of cardinality \(pq\).
Let \(G\) be a group of cardinality \(pq\) where \(p\) and \(q\) are distinct primes and \(p< q\text{.}\)
Proof.
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Since \(n_q(G)\mid p\text{,}\) we have \(n_q(G)\in \{1, p\}\text{.}\) Since \(n_q(G)\equiv 1 \pmod q\) and since \(p\not\equiv 1\pmod q\) (since \(1< p < q\)), we conclude that \(n_q(G)=1\) and thus that \(G\) has a normal \(q\)-Sylow subgroup.
Theorem 1.22.3. Groups of cardinality \(12\).
If \(G\) is a group of cardinality \(12\text{,}\) then \(G\) either \(G\) has a normal \(3\)-Sylow subgroup or \(G\cong A_4\text{.}\)
Proof.
We show that if \(G\) does not have a normal \(3\)-Sylow subgroup, then \(G\cong A_4\text{.}\) Assume \(G\) does not have a normal \(3\)-Sylow subgroup: or equivalently, \(n_3(G)\ne 1\text{.}\) Since \(n_3(G)\mid 4\) and \(n_3(G)\equiv 1 \pmod 3\text{,}\) it follows that \(n_3(G)=4\text{.}\) Let
\begin{equation*}
A=\Syl_3(G)=\{H_1, H_2,H_3,H_4\}\text{.}
\end{equation*}
By Sylow 2
Theorem 1.22.4. Groups of cardinality \(p^2q\).
Let \(G\) be a finite group of cardinality \(p^2q\text{,}\) where \(p\) and \(q\) are distinct primes.
-
If \(q> p\text{,}\) then either \(p=2\) and \(q=3\text{,}\) or \(G\) has a normal \(q\)-Sylow subgroup.
In particular, \(G\) contains a normal Sylow subgroup.
Proof.
First observe that if \(p> q\text{,}\) then since \(n_p(G)\mid q\) and \(n_p(G)\equiv 1\pmod p\text{,}\) we must have \(n_p(G)=1\text{,}\) and thus \(G\) has a normal \(p\)-Sylow subgroup.
Now assume \(p< q\) and \(n_q(G)\ne 1\text{.}\) Since \(n_q(G)\mid p\text{,}\) it follows that \(n_q(G)=p\text{,}\) and furthermore that \(p\equiv 1\pmod q\text{.}\) We have the following options for the number of Sylow subgroups:
\begin{align*}
n_p(G) \amp \in \{1,q\}\\
n_q(G) \amp \in \{1,p,p^2\}\text{.}
\end{align*}
We will show that if \(n_p(G)=q\text{,}\) then \(n_q(G)=1\text{.}\)
Example 1.22.5. Groups of cardinality 20.
Let \(G\) be a group of cardinality \(20\text{.}\)
-
Show that \(G\) is a abelian if and only \(G\) has a normal \(2\)-Sylow subgroup, and that in this case \(G\) isomorphic either to \(\Z/4Z\times \Z/5\Z\) or \(\Z/2\times \Z/2\Z\times \Z/5\Z\text{.}\)
-
A \(2\)-Sylow subgroup of \(G\) is isomorphic either to \(\Z/4\Z\) or \(\Z/2\Z\times \Z/2\Z\text{.}\) Give examples of nonabelian groups \(G\) for each case.
Solution.
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The previous theorem guarantees that \(n_5(G)=1\) and hence that \(G\) has a normal \(5\)-Sylow subgroup. You can also see this easily using the Sylow theorems since \(n_5(G)\mid 4\) implies \(n_5(G)\in \{1,2,4\}\) and \(n_5(G)\equiv 1 \pmod 5\) implies \(n_5(G)\notin\{2,4\}\text{.}\)
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The reverse implication is easy: \(G\) has a \(2\)-Sylow subgroup by Sylow 1, and this subgroup is normal since \(G\) is abelian!Now assume \(G\) has a normal \(2\)-Sylow subgroup \(N_1\text{,}\) \(\abs{N_1}=4\text{.}\) Since \(G\) has a normal \(5\)-Sylow subgroup \(N_2\text{,}\) \(\abs{N_2}=5\text{,}\) we have \(N_1\cap N_2=\{e\}\) and \(N_1N_2=G\text{.}\) It follows from a homework exercise that \(G\cong N_1\times N_2\text{.}\) Since \(N_2\cong \Z/5\Z\) and \(N_1\) is isomorphic either to \(\Z/4\Z\) or \(V\cong \Z/2\Z\times \Z/2\Z\) (the only isomorphism types of groups of cardinality \(4\)), we have\begin{equation*} G\cong \Z/4\Z\times \Z/5\Z \text{ or } G\cong \Z/2\Z\times \Z/2\Z\times \Z/5\Z\text{.} \end{equation*}In particular, \(G\) is abelian!
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The group \(D_{10}=\angvec{r,s\mid r^5=s^2=e, sr=r^{-1}s}\) is a nonabelian group whose \(2\)-Sylow subgroups are all isomorphic to \(V_4\text{.}\) Why? Because \(D_{10}\) has a cardinality-4 subgroup (Sylow 1), and \(D_{10}\) does not contain any element of order \(4\text{.}\) More explicitly, you can show that the subgroups\begin{equation*} H_1=\angvec{r^5,s}, H_2=\angvec{r^5, r^2s}, H_3=\angvec{r^5, r^4s}, H_4=\angvec{r^5, r^6s}, H_5=\angvec{r^5, r^8s} \end{equation*}are the five \(2\)-Sylow subgroups of \(D_{10}\text{.}\)The group\begin{equation*} \Aff(\Z/5\Z)=\{f\colon \Z/5\Z\to \Z/5\Z\mid f(x)=ax+b, (a,b)\in (\Z/n/Z)^*\times \Z/n\Z\} \end{equation*}is a nonabelian group of cardinality \(20\) whose \(2\)-Sylow subgroups are each isomorphic to \(V_4\text{.}\) (See definition below.) This is easiest to see by using the isomorphism\begin{equation*} \Aff(\Z/5\Z)=\left\{A\in \GL_2(\Z/5\Z)\mid A=\begin{bmatrix} a\amp b\\ 0\amp 1 \end{bmatrix}, a\in (\Z/5\Z)^*\}\leq \GL_2(\Z/5\Z)\text{.} \end{equation*}Indeed, the matrix\begin{equation*} A=\begin{bmatrix} 2\amp 0 \\ 0 \amp 1 \end{bmatrix} \end{equation*}is easily seen to have order \(4\text{,}\) and thus \(\angvec{A}\cong \Z/4\Z\text{.}\)
The last example introduced an affine transformation group \(\Aff(\Z/5\Z)\text{.}\) We give an official definition of these groups in Specimen 15. There are a number of easy claims included in that group definition that we leave as exercises to the reader.
Specimen 15. Group of invertible affine transformations.
Let \(n\geq 2\) be an integer. An affine transformation of \(\Z/n\Z\) is a function \(f\colon \Z/n\Z\rightarrow \Z/n\Z\) of the form \(f(x)=ax+b\text{,}\) where \(a,b\in \Z/n\Z\text{.}\) Such a function is invertible if and only if \(a\in (\Z/n\Z)^*\) and we denote by \(\Aff(\Z/n\Z)\) the set of all invertible affine transformations of \(\Z/n\Z\text{:}\) i.e.,
\begin{equation}
\Aff(\Z/n\Z)=\{f\colon \Z/n\Z\rightarrow \Z/n\Z\mid f(x)=ax+b, (a,b)\in (\Z/n/Z)^*\times \Z/n\Z\}\text{.}\tag{1.22.1}
\end{equation}
This set forms a group under composition of functions, and we have
\begin{equation*}
\Aff(\Z/n\Z)\cong \{A\in \GL_2(\Z/n\Z)\mid A=\begin{bmatrix}
a\amp b \\ 0 \amp 1
\end{bmatrix}, a\in (\Z/n\Z)^*\}\text{,}
\end{equation*}
as witnessed by the map
\begin{equation*}
f(x)=ax+b\longmapsto A=\begin{bmatrix}
a\amp b \\ 0 \amp 1
\end{bmatrix}\text{.}
\end{equation*}
Theorem 1.22.6. Simplicity of groups of cardinality 60.
Theorem 1.22.7. Unique simple group of cardinality 60.
Corollary 1.22.8. Group of rotational symmetries of the dodecahedron.
Let \(D\) be a regular dodecahedron in \(\R^3\text{,}\) and let \(G\) be its group of rotational symmetries. We have \(G\cong A_5\text{.}\)
Proof.
The regular dodecahedron has \(12\) pentagonal faces, \(30\) edges, and \(20\) vertices. Consider the action of \(G\) on the set \(F\) of faces of \(D\text{.}\) It is easy to see that this action is transitive, and that the stabilizer \(G_F\) of any given face \(F\) consists of the rotations of degree \(2\pi k/5\) about the axis passing through the centers of that face and its opposite face. It follows from the orbit-stabilizer theorem that
\begin{equation*}
\abs{G}=\abs{O_{F}}\abs{G_F}=12\cdot 5=60\text{.}
\end{equation*}
Furthermore, since \(G\) contains such rotations of order \(5\) for each of the 6 pairs of opposite faces, we see that it has at least \(24\) elements of order \(5\text{.}\) It follows that \(n_5(G)\ne 1\text{.}\) We conclude from Theorem 1.22.6 that \(G\) is simple, and from Theorem 1.22.7 that \(G\cong A_5\text{.}\)
