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 $p$.

If the order of a group $G$ is divisible by $p^m$ but by no higher power of $p$ for some prime $p$ then any subgroup of $G$ of order $p^m$ is called a Sylow group corresponding to $p$.

Theorem: Every group $G$ possesses at least one Sylow group corresponding to 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 $h_i \gt 1$ and $(h_i,p)=1$. Then $|N_i|$ is less than $|G|$ and divisible by $p^m$. By inductive hypthoesis, $N_i$ 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 $\mu \gt 0$. 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.

Theorem: All Sylow groups belonging to the same prime are conjugates.

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

\[G = A g_1 B \cup ... \cup A g_r B\]

and

\[ |G| = |A|B| / d_1 + ... + |A||B| / d_r \]

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

\[ r = \frac{p^m}{d_1} + ... \frac{p^m}{d_r} \]

Now $D_i$ is a subgroup of $B$, 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.

Theorem: If there are exactly $k$ Sylow groups of a group $G$ corresponding to a prime $p$ 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

\[ G = A g_1 N \cup ... \cup A g_r N \]

Then

\[ |G| = \frac{|N| p^m}{d_1} + ... + \frac{|N|p^m}{d_r} \]

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

\[ k = 1 + \frac{p^m}{d_2} +...+\frac{p^m}{d_r} \]

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 group of $G$ corresponding to $p$. 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 clearly 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 $0 \lt \mu \lt m$.

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 $m \gt 2$. 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

\[ G \triangleright A_1 \triangleright A_2 \triangleright ... \triangleright A_{m-1} \triangleright \{1\} \]

Example: There is no simple group of order 200. For let $G$ be a group with order 200. Then since $200 = 5^2 \times 8$, $G$ contains $k$ Sylow groups of order 25 where $k = 1 \bmod 5$ and $k | 200$. Thus $k | 8$ which is impossible unless $k = 1$. Thus there exists a unique normal Sylow group of order 25, and hence the group is not stimple.

Example: There is no simple group of order 30. Suppose there is such a group. Then none of its Sylow groups are unique, implying it has $1 + 5 = 6$ Sylow groups of order 5, hence there are $6\times 4 = 24$ elements of order $5$, and similarly we must have $1 + 3\times 3 =10$ Sylow groups of order 3, 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-subgroup then $G$ is simple.

Proof: Suppose $|G| = 60$ and contains more than one Sylow 5-subgroup, but there exists a proper normal subgroup. Then note we must have exactly $6$ Sylow 5-subgroups. 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 $5$-subgroup 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-subgroup, 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-subgroups.

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

TODO: proof