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Abstract algebra A Kursobjekt for Math 331

Aaron Greicius

Section 1.20 Sylow theorems: intro

Subsection 1.20.1 Stamement and illustration of Sylow theorems

Definition 1.20.1. Order at \(p\).

Let \(n\) be nonzero integer, and let \(p\) be a prime integer. The \(p\)-adic valuation of \(n\text{,}\) denoted \(v_p(n)\text{,}\) is the largest integer \(k\) such that \(p^k\mid n\text{.}\)
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Remark 1.20.2. \(p\)-adic valuation.

It follows from the fundamental theorem of arithmetic that the \(p\)-adic valuation of an integer \(n\) can be read directly from the prime factorization of \(n\text{.}\) For example, from $12=2^2\cdot 3$ we conclude
\begin{align*} v_2(12) \amp =2\\ v_3(12) \amp =1\\ v_p(12) \amp =0 \text{ for all primes } p\geq 5\text{.} \end{align*}
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Definition 1.20.3. Sylow subgroups.

Let \(G\) be a finite group of cardinality \(n\text{.}\) Let \(p\) be a prime divisor of \(n\text{,}\) and let \(v_p(n)=k\text{.}\) A \(p\)-Sylow subgroup of \(G\) is a subgroup of \(G\) of cardinality \(p^k\text{.}\) We denote the set of all \(p\)-Sylow subgroups \(\Syl_p(G)\text{:}\) i.e.,
\begin{equation*} \Syl_p(G)=\{H\leq G\mid \abs{H}=p^{k}\}\text{.} \end{equation*}
Lastly we let \(n_p(G)\) be the number of \(p\)-Sylow subgroups of \(G\text{:}\)
\begin{equation*} n_p(G)=\abs{\Syl_p(G)}\text{.} \end{equation*}
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Let’s illustrate the definitions above with some groups whose lattices of subgroups we know a great deal about already.
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Example 1.20.4. Sylow subgroups of \(S_3\).

We have \(\abs{S_3}=6=2\cdot 3\text{.}\) From the lattice of subgroups of \(S_3\text{,}\) we see immediately that
\begin{align*} \Syl_2(S_3) \amp =\{\angvec{(12)}, \angvec{(13)}, \angvec{(23)}\}\\ \Syl_3(S_3) \amp = \{\angvec{(123)}\} \end{align*}
and thus
\begin{align*} n_2(S_3) \amp = 3 \amp n_3(S_3)\amp =1 \end{align*}
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Example 1.20.5. Sylow subgroups of cyclic groups.

Let \(G=\angvec{g}\) be a finite cyclic group of cardinality \(n=\ord g\text{.}\) Fix a prime divisor \(p\) of \(n\text{,}\) and let \(v_p(n)=k\text{,}\) so that \(n=p^km\text{,}\) where \(p\nmid m\text{.}\) From Theorem 1.9.7, we know there is exactly one subgroup of \(G\) of cardinality \(p^k\text{:}\) namely, \(H=\angvec{g^{n/p^k}}=\angvec{g^m}\text{.}\) Thus
\begin{equation*} \Syl_p(G)=\{\angvec{g^m}\} \end{equation*}
and \(n_p(G)=1\text{.}\)
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Example 1.20.6. Sylow subgroups of \(A_4\).

We have \(\abs{A_4}=12=2^2\cdot 3\text{.}\) Having computed the complete lattice of subgroups of \(A_4\) (see Figure 1.16.7) we see that
\begin{align*} \Syl_2(A_4) \amp =\{\angvec{(12)(34),(13)(24)}\}\\ \Syl_3(A_4) \amp = \{\angvec{(123)}, \angvec{(124)}, \angvec{134}, \angvec{(234)}\} \end{align*}
and thus
\begin{align*} n_2(A_4) \amp = 1 \amp n_3(A_4)\amp = 4\text{.} \end{align*}
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Example 1.20.7. Sylow subgroups of \(S_4\).

We have \(\abs{S_4}=24=2^3\cdot 3\text{.}\) We have not computed the full lattice of subgroups of \(S_4\) yet, so we are not quite in a position to say immediately what the various Sylow subgroups of \(S_4\) are.
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However, since a 3-Sylow subgroup is just a group of cardinality 3 in this case, it must be a cyclic subgroup generated by a 3-cycle. Thus, we have
\begin{align*} \Syl_3(S_4) \amp = \{\angvec{(123)}, \angvec{(124)}, \angvec{134}, \angvec{(234)}\}\text{,} \end{align*}
just as in Example 1.20.6.
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What about the \(2\)-Sylow subgroups? Recall that you showed in a homework exercise that there are exactly three subgroups of \(S_4\) that are isomorphic to \(D_4\text{:}\) call them \(K_1, K_2, K_3\text{.}\) (In fact, each \(K_i\) can be described as \(\angvec{(abcd),(ac)}\text{.}\)) Thus we have
\begin{equation*} \Syl_2(S_4)\supseteq \{K_1,K_2,K_3\}\text{.} \end{equation*}
It turns out that this inclusion is in fact an equality, as we will be able to show using Sylow 2.
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Before stating the three Sylow subgroups, we make official the notion of a \(p\)-group.
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Definition 1.20.8. \(p\)-group.

Let \(p\) be a prime integer. A \(p\)-group is a group of cardinality \(p^k\) for some positive integer \(k\text{.}\) Given a group \(G\text{,}\) a \(p\)-subgroup of \(G\) is a subgroup \(H\leq G\) that is a \(p\)-group.
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The corollary below follows almost immediately from Theorem 1.20.10 by taking \(H\) to be a \(p\)-Sylow subgroup.
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The statements of Sylow 1 and Sylow 3 are easily verified in examples 1.20.4–1.20.6. Using Sylow 2, we can now verify that our description of the \(2\)-Sylow subgroups of \(S_4\) in Example 1.20.7 is complete. Why? We know that \(S_4\) has exactly three subgroups isomorphic to \(D_4\text{.}\) Since their cardinality is \(8\text{,}\) these are \(2\)-Sylow subgroups. Are there any others? No! By Sylow 2 all \(2\)-Sylow subgroups of \(S_4\) are conjugate. Since conjugation defines an isomorphism, it follows that all \(2\)-Sylow subgroups of \(S_4\) are isomorphic to \(D_4\text{.}\) Thus the three we provided are the only ones.
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In a similar vein, Sylow 2 tells us that all of the \(3\)-Sylow subgroups of \(A_4\) are conjugate. This is not immediately obvious, especially since the 3-cycles of \(A_4\) break up into two distinct conjugacy classes:
\begin{align*} C_{(123)} \amp =\{(123),(134),(243),(142)\}\\ C_{(132)} \amp =\{(132),(143),(234), (124)\}\text{.} \end{align*}
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Subsection 1.20.2 First action: \(G\) on \(p^k\)-subsets of \(G\)

Our proof of the Sylow theorems proceeds as a sequence of three well-chosen group actions, each of whose decomposition into orbits yields useful information. For the first action, we have \(G\) acting on the set \(A\) of all subsets of \(G\) of cardinality \(p^k\text{,}\) where \(v_p(\abs{G})=k\text{:}\) i.e.,
\begin{equation*} A=\{X\subseteq G\mid \abs{X}=p^k\}\text{,} \end{equation*}
the set of all \(p^k\)-subsets of \(G\text{.}\) The action here is by left multiplication: i.e., given \(X=\{g_1,g_2,\dots, g_{p^k}\}\) and \(g\in G\text{,}\) we define
\begin{equation*} g\cdot X=\{gg_1,gg_2,\dots, gg_{p^k}\}\text{.} \end{equation*}
A detailed analysis of this action is given in Proposition 1.20.14. This reveals that the \(p\)-Sylow subgroups of \(G\) are in 1-1 correspondence with the orbits of this action that are generated by an actual subgroup. Before getting into the details of that proof, it pays to look at a concrete example.
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Example 1.20.13. Action of \(D_3\) on \(2\)-subsets.

Let \(G=D_3\text{,}\) and let \(A\) be the set of all \(2\)-subsets of \(D_3\text{.}\) The action of \(G\) on \(A\) gives rise to the following orbits:
\begin{align*} O_1 \amp = \{\{e,s\},\{r,rs\},\{r^2,r^2\}\}\\ O_2 \amp = \{\{e,rs\}, \{r, r^2s\}, \{r^2,s\}\}\\ O_3 \amp = \{\{e,r^2s\},\{r,s\},\{r^2,rs\}\}\\ O_4 \amp = \{\{e,r\},\{r,r^2\},\{e, r^2\}, \{s,r^2s\}, \{s,rs\}, \{r^2s, rs\}\}\text{.} \end{align*}
Note that the three \(2\)-Sylow subgroups of \(D_3\) are contained in the three distinct orbits of size \(6/2=3\text{,}\) and that the one orbit whose cardinality is greater than \(2\) includes no \(2\)-Sylow subgroups.
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Proof.

  1. It suffices to prove that any orbit \(O_X\subseteq A\) contains a subset \(X'\) that contains \(e\text{.}\) This is easy, if \(X=\{g_1,g_2,\dots, g_{p^k}\}\text{,}\) then
    \begin{equation*} g_1^{-1}\cdot X=\{1, g_1^{-1}g_2,\dots, g_1^{-1}g_{p^k}\}\in O_X\text{.} \end{equation*}
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  2. With \(X_i\) as above, we have \(G_{X_i}\subseteq X_i\text{.}\) This is because if \(g\in G_{X_i}\text{,}\) then since \(e\in X_i\) and \(g\cdot X_i=X_i\text{,}\) we must have \(g=ge\in X_i\text{.}\) As a result, we have
    \begin{equation*} \abs{G_{X_i}}\leq \abs{X_i}=p^k\text{.} \end{equation*}
    It follows that
    \begin{equation*} \abs{O_{X_i}}=\abs{G}/\abs{G_{X_i}}\geq n/p_k=m.\text{.} \end{equation*}
    For the second statement notes that since \(\abs{O_{X_i}}\) is a divisor of \(n=p^km\text{,}\) it must be of the form \(p^im\) or \(p^i d\) for some \(d\mid m\) with \(1\leq d< m\text{.}\) From this it follows that \(\abs{O_{X_i}}> m\) if and only if \(p\mid \abs{O_{X_i}}\text{.}\)
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  3. Since \(\abs{O_{X_i}}\mid n=p^km\) and \(\abs{O_{X_i}}\geq m\text{,}\) it follows that \(\abs{O_{X_i}}=p^{n_i}m\) for some \(0\leq n_i\leq k\text{.}\)We have \(\abs{O_{X_i}}=m\) if and only if \(\abs{G_{X_i}}=p^k\) if and only if \(G_{X_i}=X_i\text{,}\) since \(G_{X_i}\subseteq X_i\) and \(\abs{G_{X_i}}=\abs{X_i}=p^k\text{.}\) In this case \(X_i=G_{X_i}=H_i\) is a \(p\)-Sylow subgroup of \(G\text{,}\) and by definition of the group action, the orbit \(O_{H_i}\) is literally the coset space \(G/H_i\text{.}\)
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  4. We claim that the \(p\)-Sylow subgroups of \(G\) are in 1-1 correspondence with the orbits of \(A\) of cardinality \(m\text{.}\) Indeed, since each such orbit is of the form \(G/H_i\) for some \(H_i\in \Syl_p(G)\text{,}\) and since \(H_i\) is the only subgroup of the coset space \(G/H_i\text{,}\) we see that each such orbit contains exactly one \(p\)-Sylow subgroup. Furthermore, for any \(p\)-Sylow subgroup \(H\) its orbit \(G_H\) must have cardinality \(n/p^k=m\text{.}\)
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    Let \(O_{X_1}, O_{X_2},\dots, O_{X_s}\) be the distinct orbits of \(A\) of cardinality \(m\text{.}\) We have
    \begin{align*} {n\choose p^k} \amp =\sum_{i=1}^r\abs{O_{X_i}}\\ \amp =\sum_{i=1}^s\abs{O_{X_i}}+\sum_{i=s+1}^r\abs{O_{X_i}}\\ \amp =m\cdot n_p(G)+\sum_{i=s+1}^rp^{n_i}m\text{,} \end{align*}
    where \(n_i\geq 1\) for all \(s+1\leq i\leq r\text{.}\) The last line follows since there are exactly \(n_p(G)\) orbits of cardinality \(m\text{,}\) and since all other orbits have cardinality \(p^jm\) for some \(j\geq 1\text{.}\) As a result, we have
    \begin{align*} {n\choose p^k} \amp \equiv m\cdot n_p(G)\text{.} \end{align*}
    Now for a wonderful trick. Since the equivalence above holds for any group of cardinality \(n\text{,}\) in particular it holds for a cyclic group. But in this special case we have \(n_p(G)=1\text{.}\) It follows that
    \begin{equation*} {n\choose p^k}\equiv m \pmod p\text{.} \end{equation*}
    Lastly, since \(\gcd(m,p)=1\text{,}\) it is invertible modulo \(p\text{,}\) allowing us to conclude that
    \begin{equation*} 1\equiv n_p(G) \pmod p\text{,} \end{equation*}
    as desired.
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Note that Proposition 1.20.14 includes the formula (1.20.2) among its results, giving us a partial proof of Sylow 3. Additionally, formula (1.20.2) implies Sylow 1, as we me now make official.
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Proof.

Since \(n_p(G)\equiv 1\pmod p\) and \(0\not\equiv 1\pmod p\text{,}\) it follows that \(n_p(G)\ne 0\text{.}\) Equivalently, \(\Syl_p(G)\ne \emptyset\text{.}\)
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Before leaving this first group action in our suite of group actions, we extract from the proof of Proposition 1.20.14 a wonderful gem of a combinatorial fact.
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