Sylow Groups

Lemma: Let \(A\) be an abelian group. If \(p\) is a prime factor of \(|A|\) then \(A\) contains at least one element of order \(p\).

Proof: The lemma is trivial when \(|A| = p\), which we shall use to start an induction. Assume \(|A|\) is composite. Then \(A\) contains a proper subgroup. Choose a proper subgroup \(H\) of maximum order. If \(p | |H|\) then by induction \(H\) contains an element of order \(p\), so assume \((|H|,p) = 1\). Then take some element \(g \in A \setminus H\). Let \(t\) be the order of \(g\). Consider the group \(A' = H \langle g \rangle\). Since \(A\) is abelian, we have \(H \langle g \rangle = \langle g \rangle H\) thus \(A'\) is a group. But since it strictly contains \(H\), we have \(A' = A\) by maximality of \(H\).

Now \(H \langle g \rangle\) contains \(|H|t / d\) elements where \(d = |H \cap \langle g \rangle|\). Thus \(|A| d = |H| t\). Since the \(p\) divides the left-hand side, and \((|H|,p) = 1\), we must have \(p | t\), and \(g^{t / p}\) is an element of order \(p\).

Let \(p^m\) be the highest prime power dividing the order of a group \(G\) for some prime \(p\). Any subgroup of \(G\) of order \(p^m\) is called a Sylow \(p\)-group.

Sylow’s First Theorem: Every group \(G\) possesses at least one Sylow \(p\)-group for each prime factor of \(|G|\).

Proof: The theorem is immediate when \(|G| = 2\), which we shall use to start an induction. Write \(|G| = p^m r\) where \((r, p) = 1\). Decompose \(G\) into classes of conjugate elements, and pick elements \(a_1, ..., a_k\) from each class. Recall if \(h_i\) denotes the size of the class containing \(a_i\) we have \(|G| = h_1 + ... + h_k\). Also recall the normalizer \(N_i\) of \(a_i\) satisfies \(|N_i| = |G|/h_i\). We have two cases:

Case 1: Suppose there exists \(h_i\) with \(h_i > 1\) and \((h_i,p)=1\). Then \(|N_i|\) is less than \(|G|\) and divisible by \(p^m\). By inductive hypothesis, \(N_i\) possesses a subgroup of order \(p^m\) which is the Sylow group corresponding to \(p\).

Case 2: For all \(i\), we have \(h_i = 1\) or \(p | h_i\). We have \(h_i = 1\) for self-conjugate elements, and we must have at least one of these since \(1\) is self-conjugate. Let \(z\) be the number of self-conjugate elements. Then \(p^m r = z + x p\) for some integer \(x\), hence \(p | z\). Thus the order of the center is divisible by \(p\). Since it is abelian, by the lemma it contains at least one element \(g\) that commutes with all elements and has order \(p\). Then \(P = \langle g \rangle\) is a normal subgroup of \(G\) and \(G / P\) has order \(p^{m-1} r\). By the inductive hypothesis \(G/P\) contains a Sylow group of order \(p^{m-1}\), which we write \(H/P\) where \(H\) is a subgroup of \(G\). Then \(p^{m-1} = |H|/p\), thus \(|H| = p^m\) and \(H\) is a Sylow group of \(G\) corresponding to \(p\).

Theorem: [Cauchy] Let \(G\) be a group. If \(p\) is a prime factor of \(|G|\) then \(G\) contains at least one element of order \(p\)

Proof: Let \(H\) be a Sylow group of \(G\) of order \(p^m\). If \(1 \ne h\in H\) then the order of \(h\) is \(p^\mu\) for some \(\mu > 0\). Then \(h^{p^{\mu - 1}}\) has order \(p\).

All subgroups conjugate to a Sylow group are themselves Sylow groups. It turns out the converse is true.

Sylow’s Second Theorem: All Sylow \(p\)-groups for a prime \(p\) are conjugates.

Proof: Let \(A, B\) be subgroups of \(G\) of order \(p^m\). Recall we can decompose \(G\) relative to \(A\) and \(B\):

and

where \(d_i\) is the size of \(D_i = g_i^{-1} A g_i \cap B\). We have \(|A| = |B| = p^m\) and \(|G| = p^m r\) where \((r,p)=1\). Thus dividing by \(p^m\) gives

Now \(D_i \le B\), hence \(d_i\) is some nonnegative power of \(p\) and is at most \(p^m\). Since \((r,p) = 1\), we must have \(p^m / d_l = 1\) for some \(l\), in other words \(d_l = p^m\). Then \(D_l\) has the same order as \(B\) and is contained in \(B\), thus \(D_l = B\) and similarly \(D_l = g_l^{-1} A g_l\). Hence \(B = g_l^{-1} A g_l\) implying that \(A, B\) are conjugate.

Corollary: A Sylow group is unique if and only if it is a normal subgroup.

Sylow’s Third Theorem: Let \(k\) be the number of Sylow \(p\)-groups of a group \(G\). Then \(k = 1 \bmod p\) and \(k\) divides \(|G|\).

Proof: We know that the number of distinct Sylow groups is equal to the number \(k\) of distinct conjugates. Let \(A\) be some Sylow group corresponding to \(p\) and let \(N\) be the normalizer of \(A\). Recall \(|G| = |N| k\) thus \(k\) divides \(|G|\).

Every \(a\in A\) satisfies \(a^{-1} A a = A\) thus \(a \in N\), Hence \(A\triangleleft N\). Thus \(|N| = p^m n'\) where \((n',p)=1\).

Decompose \(G\) as the disjoint sets

Then

where \(d_i\) is the order of the group \(D_i = g^{-1}_i A g_i \cap N\). Without loss of generality assume \(g_1 = 1\), hence \(A g_1 N = A N = N\). Now dividing by \(n\) gives

Now suppose \(d_i = p^m\) for some \(i\). Then \(D_i = g_i^{-1} A g_i\), implying \(g_i^{-1} A g_i \subset N\). Now \(N\) possesses a Sylow group of order \(p^m\), and we have already found two: \(A, g_i^{-1} A g_i\). But \(A\) is normal in \(N\) thus must be the unique Sylow group, hence \(A = g_i^{-1} A g_i\). Since \(N\) is the normalizer of \(A\) we must have \(g_i \in N\) and hence \(A g_i N = A N = N\), which is impossible unless \(i = 1\).

Thus all terms in the above summation are divisble by \(p\) except for the first term which is equal to one.

Theorem: Any group \(G\) of order \(p q\) for primes \(p,q\) satisfying \(p \ne 1 \pmod{q}\) and \(q \ne 1 \pmod{p}\) is abelian.

Proof: We have already shown this for \(p = q\) so assume \((p,q)=1\). Let \(P = \langle a \rangle\) be a Sylow \(p\)-group of \(G\). The number of such subgroups is a divisor of \(p q\) and also equal to \(1\) modulo \(p\). Also \(q \ne 1 \bmod p\). Then since the number of such subgroups cannot be equal to \(p, q, p q\), it must be equal to one. By the above corollary we have that \(P\) is normal in \(G\) of order \(p\). Similarly we can find a group \(Q = \langle b \rangle\) normal in \(G\) of order \(q\).

Then \(P Q = Q P\), which by the product theorem is a subgroup order \(p q / |P \cap Q|\). But since \((p,q)=1\) they only have the identity element in common thus \(G = P Q\). Also, recall these conditions also imply every element of \(P\) commutes with every element of \(Q\). Then every element of \(G\) has the form \(a^\alpha b^\beta = b^\beta a^\alpha\) and is abelian.

A prime power group is a group whose order is a power of a prime. [It seems that nowadays they are referred to as \(p\)-groups.] All Sylow groups are prime power groups. Recall that a group \(G\) of order \(p^m\) for a prime \(p\) has at least one nontrivial self-conjugate element, thus we can find a self-conjugate element of order \(p\). Let \(a\) be such an element. Then \(x^{-1} a x\) for any \(x \in G\), and \(\langle a \rangle\) is a normal subgroup of order \(p\). In general:

Theorem: A group of order \(p^m\) for a prime \(p\) contains at least one normal subgroup of order \(p^\mu\) for any \(0 < \mu < m\).

Proof: The theorem is true for \(m=2\) because in this case the group is abelian. We shall use this case to base an induction.

Suppose \(G\) is a group of order \(p^m\) for \(m > 2\). Then let \(P\) be a normal subgroup of \(G\) of order \(p\). Then \(G/P\) has order \(p^{m-1}\) which by inductive assumption has an invariant subgroup of order \(p^{\mu - 1}\) which has the form \(A/P\) for some normal subgroup \(A\) in \(G\) with order \(p^\mu\).

Corollary: All prime power groups are soluble.

Proof: A group \(G\) of order \(p^m\) has a normal subgroup \(A_1\) of order \(p^{m-1}\) which in turn contains a normal subgroup of order \(p^{m-2}\), and so on. Thus we can construct the composition series

Example: There is no simple group of order 200.

Let \(G\) be a group of order \(200 = 5^2 \times 8\) with \(k\) Sylow 5-groups of order 25. Then \(k = 1 \bmod 5\) and \(k | 200\). Thus \(k | 8\) which forces \(k = 1\). Hence there exists a unique normal Sylow 5-group of order 25, and the group is not simple.

Example: There is no simple group of order 30.

Suppose there is such a group. Then none of its Sylow groups are unique, implying \(1 + 5 = 6\) Sylow 5-groups, hence there are \(6\times 4 = 24\) elements of order \(5\), and similarly \(1 + 3\times 3 = 10\) Sylow 3-groups, thus the total number of elements is greater than 30, a contradiction.

We can now supply an alternative proof that \(A_n\) is simple for \(n\ge 5\):

Proposition: If \(|G| = 60\) and \(G\) has more than one Sylow 5-group then \(G\) is simple.

Proof: Suppose \(|G| = 60\) and contains more than one Sylow 5-group, but there exists a proper normal subgroup. Then we must have exactly 6 Sylow 5-groups. Let \(P\) be such a group. Then the normalizer of \(P\) has order 10 since its index is \(6\).

If \(5 | |H|\) then \(H\) contains a Sylow 5-group of \(G\) and since \(H\) is normal it contains all \(6\) conjugates of this subgroup, hence \(|H| \ge 1 + 6\cdot 4 = 25\) hence we must have \(|H|=30\). But by the previous example, \(|G|\) must have a unique Sylow 5-group, a contradiction, thus 5 does not divide \(|H|\).

If \(|H|\) is 6 or 12 then \(H\) has a normal Sylow subgroup of order 2,3, or 4, which is also normal in \(G\), and we may replace \(H\) by this. Hence \(|G/H| = 30, 20\) or \(15\). Then by previous results, \(G/H\) has a normal subgroup of order \(5\). Its preimage under the natural map is a normal subgroup whose order is a multiple of 5, which we have previously shown to be a contradiction.

Corollary: \(A_5\) is simple.

Proof: The subgroups \(\langle (1 2 3 4 5) \rangle\) and \(\langle (1 3 2 4 5) \rangle\) are distinct Sylow 5-groups.

Theorem: \(A_n\) is simple for all \(n\ge 5\).

Proof: Let \(n \ge 6\). Suppose \(H\) is a proper normal subgroup of \(A_n\), and let \(1 \ne \tau \in H\).

For \(i \in [1..n]\) let \(G_i\) be the permutations that leave \(i\) fixed in place. Then \(G_i \le A_n\) and \(G_i\) is isomorphic to \(A_{n-1}\). By inductive assumption, each \(G_i\) is simple.

First suppose there exists an \(i\) such that \(\tau(i) = i\). Then \(H \cap G_i\) is non-trivial and a normal subgroup of \(G_i\). Since \(G_i\) is simple, we have \(H \cap G_i = G_i\), thus \(G_i\) is a subgroup of \(H\).

For any \(\sigma \in A_n\), since \(\tau(x) = y\) implies \(\sigma \tau \sigma^{-1} (\sigma(x)) = \sigma(y)\):

The left-hand side is a subgroup of \(\sigma H \sigma^{-1} = H\), so by choosing \(\sigma\) appropriately we have \(G_j \le H\) for all \(j \in [1..n]\).

Recall any element of \(A_n\) can be written as an even number of transpositions. Since \(n > 4\), the product of any two transpositions must leave at least one item unpermuted, hence lies in \(G_j\) for some \(j\). Therefore:

implying \(A_n \le H\), a contradiction.

Thus \(\tau(i) \ne i\) for all \(i\).

Suppose the cycle decomposition of \(\tau \in H\) contains a cycle of a length 3 or more. We relabel so that:

Pick \(\sigma \in A_n\) such that \(\sigma(1) = 1, \sigma(2) = 2, \sigma(3) \ne 3\). Then:

again since \(\tau(x) = y\) implies \(\sigma \tau \sigma^{-1} (\sigma(x)) = \sigma(y)\).

Hence \(\tau(1) = \tau_1(1) = 2\), so \(\tau^{-1} \tau_1 (1) = 1\).

Since \(\tau \ne \tau_1\), we have nonidentity element \(\tau_{-1} \tau_1 \in H\) that fixes 1, a contradiction.

Thus the cycle decomposition of \(\tau\) contains only transpositions. Since \(n \ge 6\) and no element is fixed by \(\tau\) there are at least 3 transpositions. Relabel so that:

Let \(\sigma = (12)(35) \in A_n\). Then:

hence \(\tau(1) = \tau_1(1) = 2\), which as above is a contradiction.