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 $A^{\prime }=H⟨g⟩$. Since $A$ is abelian, we have $H⟨g⟩=⟨g⟩H$ thus $A^{\prime }$ is a group. But since it strictly contains $H$, we have $A^{\prime }=A$ by maximality of $H$.

Now $H⟨g⟩$ contains $|H|t/d$ elements where $d=|H\cap ⟨g⟩|$. 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 $p^{m}r=z+xp$ 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 $P=⟨g⟩$ 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 $k=1modp$ 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 $a^{-1}Aa=A$ thus $a\in N$, Hence $A◃N$. Thus $|N|=p^m{n}^{\prime }$ where $(n^{\prime },p)=1$.

Decompose $G$ as the disjoint sets

Then

where $d_i$ is the order of the group $D_i=g_i^{-1}A{g}_i\cap N$. Without loss of generality assume $g_1=1$, hence $A{g}_{1}N=AN=N$. 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 $A{g}_{i}N=AN=N$, 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 $pq$ for primes $p,q$ satisfying $p\ne 1(\mathrm{mod}q)$ and $q\ne 1(\mathrm{mod}p)$ is abelian.

Proof: We have already shown this for $p=q$ so assume $(p,q)=1$. Let $P=⟨a⟩$ be a Sylow $p$-group of $G$. The number of such subgroups is a divisor of $pq$ and also equal to $1$ modulo $p$. Also $q\ne 1modp$. Then since the number of such subgroups cannot be equal to $p,q,pq$, 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 $Q=⟨b⟩$ normal in $G$ of order $q$.

Then $PQ=QP$, which by the product theorem is a subgroup order $pq/|P\cap Q|$. But since $(p,q)=1$ they only have the identity element in common thus $G=PQ$. 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 $x^{-1}ax$ for any $x\in G$, and $⟨a⟩$ 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 $k=1mod5$ 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 $⟨(12345)⟩$ and $⟨(13245)⟩$ 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.