Useful Theorems

Theorem 0.1 (Sylow's Theorem)   <>Let ${\sl\large G}$ be a group of order $p^{\alpha}m$where (p,m) = 1.

1.

Sylow p-subgroups exist

2.

If ${\sl\large P}$ is a Sylow p-subgroup of ${\sl\large G}$ and ${\sl\large Q}$ is any p-group of ${\sl\large G}$, then there exists $g\in {\sl\large G}$ such that ${\sl\large Q} \le g{\sl\large P}g^{-1}$. In particular, any two Sylow p-subgroups of ${\sl\large G}$are conjugate in ${\sl\large G}$.

3.

The number of Sylow p-subgroups of ${\sl\large G}$ is of the form 1 + kp

$$n_p \equiv 1(mod p).$$

Further, n_p is the index in ${\sl\large G}$ of the normalizer ${\sl\large N_G(P)}$ for any ${\sl\large P}$, therefore

$$n_p \mid m$$

.

An example of how this theorem can be useful is to look at a group ${\sl\large G}$ where $\mid {\sl\large G} \mid = 14$. Sylow's Theorem says there there is a subgroup ${\sl\large P} \le Syl_{7}({\sl\large G})$. Furthermore, $n_p \mid 2$ and so n₇ = 1 and ${\sl\large P} \triangleleft {\sl\large G} \Longrightarrow {\sl\large G}$ is simple.

Theorem 0.2 (Generalized Cayley's Theorem)   <>Let ${\sl\large G}$ be a group, let ${\sl\large H}$ be a subgroup of ${\sl\large G}$ and let ${\sl\large G}$ act by left multiplication on the set ${\sl\large A}$ of left cosets of ${\sl\large H}$ in ${\sl\large G}$. Let $\pi_k$ be the associated permutation representation afforded by this action. Then

1.

${\sl\large G}$ acts transitively on ${\sl\large A}$.

2.

the stabilizer in ${\sl\large G}$ of the point $1{\sl\large H} \in {\sl\large A}$ is the subgroup ${\sl\large H}$.

3.

the kernel of $\pi_k$ is $\bigcap_{x \in {\sl\large G}} x{\sl \large H}x^{-1}$, and $\pi_k$ is the largest normal subgroup of ${\sl\large G}$ contained in ${\sl\large H}$.

proof: (of part 3)

By definition of $\pi_k$

$$\begin{eqnarray*}ker \pi_k & = & \{ g \in G \mid gx{\sl\large H} = x{\sl\large H... ...05cm] & = & \bigcap_{x \in {\sl\large G}}x{\sl\large H}x^{-1}\\ \end{eqnarray*}$$

The last part of the theorem follows from the fact that $ker \pi_h \triangleleft {\sl\large G}$ and is a subgroup of ${\sl\large H}$. If ${\sl\large N}$ is any normal subgroup of ${\sl\large G}$ in ${\sl\large H}$, then ${\sl\large N} = x{\sl\large N}x^{-1} \le x{\sl\large H}x^{-1}$ for all $x \in {\sl\large G}$ so that

$${\sl\large N} \le ker\pi_h$$

We have proved that every normal subgroup of ${\sl\large G}$ contained in ${\sl\large H}$ is also contained in $ker\pi_k$. Essentially, this means that if ${\sl\large G}$ is simple and has a subgroup of index k then ${\sl\large G}$ is isomorphic to a subgroup of S_k. If this we're not true we can find a simple contradiction. Suppose that $\mid {\sl\large G} \mid$ does not divide $\mid {\sl\large G}:{\sl\large H}\mid ! = \mid S_k \mid$. But $\mid {\sl\large G}/ker\pi_H \mid$ divides $\mid S_k \mid \Longrightarrow ker\pi_H > 1$. Therefore, $ker\pi_H$ is a non-trivial normal subgroup of ${\sl\large G}$. This is a contradiction so $\mid {\sl\large G} \mid$ divides $\mid S_k \mid$.

Theorem 0.3   <>An integer of the form 2n where n is an odd number greater than 1, is not the order of a simple group.

proof:

Let ${\sl\large G}$ be a group of order 2n for n > 1. From Cayley's Theorem, the homomorphism $\phi:g \longrightarrow \pi_G$ is an isomorphism from ${\sl\large G}$ to a permutation group on the elements of ${\sl\large G}$, where $\pi_g(x)$ is defined as gx for all $x \in {\sl\large G}$. By Sylow's Theorem ${\sl\large G}$ has an element of order 2. Pick $g\in {\sl\large G}$ with order 2. Then $\pi_g(x_i) = x_j$, $i\ne j$, and $(\pi_g)^{2}(x_i) = x_i$. Therefore we can arrange all the elements in ${\sl\large G}$ in the form

(x₀, x₁)(x₂, x₃).....(x_2n-3,x_2n-2)(x_2n-1,x_2n)

where each ordered pair is $(\pi_g(x_i),(\pi_g)^{2}(x_i))$ and ${\sl\large G} = \bigcup^{2n}_{\i=0}x_{i}$. Therefore, in cycle form, $\pi_G$ has exactly n cycles, and so $\pi_G$ is an odd permutation. This implies that the set of all even permutations in the image of ${\sl\large G}$is a normal subgroup having index 2 $\Longrightarrow {\sl\large G}$ is not simple.

Theorem 0.4   <>If ${\sl\large G}$ is a finite group of order n and p is the smallest prime dividing $\mid {\sl\large G} \mid$, then any subgroup with index p is normal in ${\sl\large G}$

proof:

Suppose ${\sl\large H} \le {\sl\large G}$ and $\mid {\sl\large G}:{\sl \large H} \mid = p$. Let ${\sl\large K}$ be the kernel of $\pi_H$ where $\pi_H$ is the permutation representation afforded by left multiplication on the left cosets of ${\sl\large H}$. Then $\mid {\sl\large G}:{\sl \large K} \mid = \mid {\sl\large G}:{\sl\large H} \mid \mid {\sl\large H}:{\sl\large K} \mid = pk$ by the Third Isomorphism Theorem. But ${\sl\large G}/{\sl\large K} \cong {\sl\large S}_p \Longrightarrow pk \mid p!$ and so $k \mid \frac{p!}{p} = (p-1)!$. But all prime divisors of (p-1)! are less than p and $k \ge p$ since it the smallest prime divisor of the groups order. Therefore k=1 and $\mid {\sl\large G}:{\sl\large K} \mid = p = \mid {\sl\large G}:{\sl\large H} \mid$, and ${\sl\large H} = {\sl\large K} \triangleleft {\sl\large G}$. ${\sl\large H} \triangleleft {\sl\large G}$, and the proof is complete.

Theorem 0.5   <>The group of even permutations ${\sl\large A}_{n}$ is simple for all $n \ge 5$.

In order to prove this we need the following result.

lemma: Let $\sigma , \tau$ be elements of the symmetric group ${\sl \large S}_{n}$ and suppose $\sigma$ has cycle decomposition

(a₁a₂....a_k)(b₁b₂....b_j)....

Then $\tau \sigma \tau^{-1}$ has cycle decomposition

$$(\tau (a_{1})(\tau (a_{2})....\tau (a_{k}))(\tau (b_{1})\tau (b_{2})....\tau (b_{k}))...,$$

proof:

We will prove by induction on n, with the assumption that ${\sl\large A}_{5}$ is simple. So assume true for all $5 \le k \le n$.

Let ${\sl\large G} = {\sl\large A}_{n}$ and suppose ${\sl\large H} \triangleleft {\sl\large G}$ where ${\sl\large H} \ne 1$ and ${\sl \large H} \ne {\sl\large G}$.

for each $i \in \{ 1,2,3,...,n \}$ let ${\sl\large G}_{i}$ be the stabilizer of i in the action of ${\sl\large G}$ on $\{1,2,3,...,n \}$. So ${\sl\large G}_{i} \le {\sl\large G}$ and note that ${\sl \large G}_{i} \cong {\sl\large A}_{n-1}$ for each i.

By induction ${\sl\large G}_{i}$ is simple.

Suppose that $\tau \in {\sl\large H}$, and that $\tau \ne 1$ and $\tau (i) = i$.

Then since ${\sl\large H} \triangleleft {\sl\large G} \Longrightarrow ({\sl\large H} \bigca... ...t {\sl\large G}_{i} \Longrightarrow {\sl\large H} \bigcap {\sl\large G}_{i} = 1$ or ${\sl\large H}$. But ${\sl\large H} \ne 1$ so

$${\sl \large G}_{i} \le {\sl\large H}$$

Therefore, ${\sl\large G}_{j} \le {\sl\large H}$ for all $j \in \{ 1,2,3,...,n \}$ and $(<{\sl\large G}_{1},{\sl\large G}_{2},....{\sl \large G}_{n}>) \le {\sl\large H}$

$$\Longrightarrow {\sl\large G} \le {\sl\large H}$$

which is a contrdiction. Therefore $({\sl\large G}_{i} \bigcap {\sl\large H}) = 1$ and $\tau(i) \ne 1$ for any $\tau \in {\sl\large H}$.
Next suppose that $\tau_{1}(i) = \tau_{2}(i)$ for some $\tau_{1}, \tau_{2} \in {\sl\large H}$. Then it follows that $(\tau_{1})^{-1} \tau_{2}(i) = i \Longrightarrow (\tau_{1})^{-1} \tau_{2} = 1 \Longrightarrow \tau_{1} = \tau_{2}$. So two elements in ${\sl\large H}$ can not take i to the same thing.

Now suppose that there is some $\tau \in {\sl\large H}$ with a cycle decomposition containing length greater than or equal to three.

$$\tau = (a_1,a_2,a_3,....)(b_1,b_2,....)...$$

Let $\sigma \in {\sl\large G}$ such that $\sigma (a_{1}) = a_{1}$ and $\sigma (a_{2}) = a_{2}$ and $\sigma (a_{3}) \ne a_{3}$. Given that ${\sl\large H} \triangleleft {\sl\large G}$ and by lemma

$$\tau_{1} = \sigma \tau \sigma^{-1} = (a_{1}, a_{2}, \sigma (a_{3}),...)(\sigma (b_{1}), \sigma (b_{2}),....)....$$

But this would imply that

$$\tau (a_{1}) = \tau_{1} (a_{1}) = a_{2}$$

which contradicts $\tau_{1}(i) \ne \tau_{2}(i)$. Therefore, only two cycles may appear in cycle decomposition of elements in ${\sl\large H}$.

So, let $\tau = (a_{1},a_{2})(a_{3},a_{4})(a_{5},a_{6})......$ and let $\sigma = (a_{1},a_{2})(a_{3},a_{5}) \in {\sl\large G}$.

$$\begin{eqnarray*}\sigma \tau \sigma^{-1} & = & (a_{1},a_{2})(a_{3},a_{5})(a_{1},... ...\ \tau_{j} & = & (a_{1},a_{2})(a_{3},a_{5})(a_{3},a_{4})....\\ \end{eqnarray*}$$

This is a contradiction since $\tau_{j}(i) \ne \tau_{k}(i)$. Therefore, the normal subgroup ${\sl\large H}$ can only have a cycle decomposition of 1-cycles. So, ${\sl\large H} = 1$ which is a contradiction. This completes the proof that ${\sl\large A}_{n}$ is simple for all $n \ge 5$.

Theorem 0.6 (Index Theorem)   <>If G is a finite group and H is a proper subgroup of G such that $\mid G \mid$ does not divide $\mid G:H \mid !$, then H contains a nontrivial normal subgroup of G. In particular, G is not simple.

proof:

Let $\phi:G \longrightarrow S_{k}$. Then the $ker \phi$ is a normal subgroup of G contained in H, and $G/ ker \phi$ is isomorphic to a subgroup of S_k. Thus, $\mid G/ ker \phi \mid$ divides $\mid S_{k} \mid = \mid G:H \mid !$. But since, $\mid G \mid$ does not divide $\mid G:H \mid !$, the order of $ker \phi$ must be greater than 1. Therefore, G is not simple.


Theorem 0.7 If P = k-1 and P is a p-cycle then $\mid N_{S_{K}}(P) \mid = p(p-1)$.


proof:

There are (p-1)! number of p-cyles in S_K. Therefore, there are (p-2)! number of Syl_P subgroups in S_k. Therefore,

$$\mid N_{S_{K}}(P) \mid = \frac{\mid S_{k} \mid }{\mid S_{P}:N_{S_{k}}(P) \mid} = p(p-1)$$

The following section will cover some useful properties of a special type of group called p-Groups. These are groups with order $p^{ \beta }$ for $\beta \ge 1$.