Quotient groups work like subgroups to obtain a “smaller” group from some base group G.
Definitions
Let \varphi be a homomorphism from G to H. The fibers of \varphi are the sets of elements of G that map to single elements of H.
For example, given the groups G = (\mathbb{Z},+) and H = Z_n be the cyclic group of order n under multiplication. We can then define the map
\varphi \colon \mathbb{Z} \to Z_n
a \rightarrowtail x^a
The fiber of \varphi over x^a is therefore
Put fiber
The homomorphism \varphi has two fibers, we will notate X_{[1]} as the fiber over [1], and X_{[0]} as the fiber over [0]. Then
X_{[1]} = \{1,3,5,\cdots\} \quad X_{[0]} = \{0,2,4,\cdots\}
The set of fibers of \varphi forms a group. We define our group operation as follows
X_{a}X_{b} = X_{ab}
Where X_{ab} is the fiber over the product ab. This is called the quotient group of \varphi. Below is a formal proof the collection of fibers forms a group.
Prove the set of fibers of \varphi is a group.
We can now look at some properties of homomorphisms
If \varphi \colon G \to H is a homomorphism, the kernal of \varphi is the set
\ker\varphi = \{g \in G \mid \varphi(g) = e_H\}
The kernal is therefore the fiber over e_H, the identity element of H.
If \varphi \colon G \to H is a homomorphism, the image of \varphi is the set
\text{im}\,\varphi = \{\varphi(g) \mid g \in G\}
The kernal and image of \varphi are subgroups of G and H respectivly.
If G,H are groups, and \varphi \colon G \to H is a homomorphism, then
\ker \varphi \leq G \: \text{ and } \: \text{im} \,\varphi \leq H
Since e_G is such that \varphi(e_G) = e_H, we know \ker \varphi \not = \varnothing. If x,y are in the \ker \varphi, then xy^{-1} is in the kernal, thus \ker \varphi is a subgroup of G.
Now take the image of \varphi. Note
\varphi(e_G) = e_H \in \text{im } \varphi \implies \text{im } \varphi \not = \varnothing
Given x,y \in \text{im } \varphi, we know that xy^{-1} is in \text{im } \varphi. Thus \text{im } \varphi \leq H.
And we can now provide a formal definition for a quotient group
Let \varphi \colon G \to H be a surjective homomorphism with \ker K. The quotient group G/K is the group whose elements are the fibers of \varphi with group operation inherited from H.
This definition requires knowing \varphi explictly. You can however define the group operation on fibers in terms of representatives.
Let \varphi \colon G \to H be a homomorphism with \ker k. Let X \in G/K be the fiber above a \in H. X = \varphi^{-1}(a). Then for any u \in X,
X = \{uk\mid k \in K\}
Let u \in X, then \varphi(u) = a. Let
uK = \{uk \mid k \in K\}
We want to show that X is equal to uK. We will first show that uK is a subset of X. For k \in K
\varphi(uk) = \varphi(u)\varphi(k) = \varphi(u) = a
Thus uK \subseteq X. Now to show that X \subseteq uK, let g \in X and k = u^{-1}g.
\varphi(k) = \varphi(u^{-1})\varphi(g) = a^{-1}a = e_H
Thus k \in \ker \varphi since k = u^{-1}g. Thus X \subseteq uK
Thus we can write any elements of the quoitent group as the set uk for all k \in K.
For any N \leq G and g \in G,
gN = \{gn\mid n \in N\}
This is a (left) coset of N in G.
Let \varphi \colon G \to H homomorphism with a kernal K. The set of cosets of K in G with operation uK \star vK = (uv)K forms a group.
Let X,Y \in G/K, and Z = XY \in G/K. Then for some a,b \in H,
X = \varphi^{-1}(a)\: Y = \varphi^{-1}(b)
This implies Z = \varphi^{-1}(ab). Let u,v be representations of X,Y respectivly. We want to show that uv \in Z. Which is only true If
\varphi(uv) \in ab \leftrightarrow \varphi(u)\varphi(v) = ab
Which is true! Thus by our previous proposition Z = uvK.
You cannot define G/N for any N \leq G.
You cannot define G/N for any N \leq G.
Let \varphi \colon G \to H be a homomorphism with Kernal K. Then
gKg^{-1} \subseteq K \; \forall g \in G
\varphi(gkg^{-1}) = e
If we have a subgroup N of $G such that gNg^{-1} \subseteq N for all g \in G, then multiplication in G/N is well defined.
G/N \times G/N \rightarrow G/N
(xN, xG)
If x_1N = x_2N, y_1N=y_2N then x_1^{-1}x_2 \in N and y_1^{-1}y_2 \in N
Orbit stabalizers
Define
b = g \cdot a \rightarrowtail gG_a
The action of G on a is called transitive if there is only one orbit
D_8 acts transitivly on each of it’s 4 vertices.
If G acts trivially on A, then G_a = G,\, \forall a \in A. Then the orbits
orb(a) = \{b \mid b = g \cdot a \text{ for } g \in G\} = a.
This is an example of a non-transitive action. If however
G = S_n
Then there always exists a g \in G such that a = g \cdot b.
Let (G,\diamond) be a group, and A = G. Define
g \cdot a = g \diamond a
then
(1) g_2 \cdot (g_1 \cdot a) = g_2 g_1 a
e\cdot a = ea = a
Let G be a group such that H \leq G
A = \{gH\mid g\in G\}
Define g \cdot aH = gaH.
Suppose A has n elements
a_1H \dots a_nH
We can describe \sigma_g as \sigma_g(i) = j iff ga_i = a_j
Let G = D_8, and H = \langle s \rangle. Then
A = \{eH,rH,r^2H,r^3H\}
Label the four elements 1,2,3,4 respectivly. We can compute \sigma_s
\begin{align*}
&s\cdot eH = sH = H\quad &\sigma_s(1) = 1\\
&s\cdot rH = srH = r^3H\quad &\sigma_s(2) = 4\\
&s\cdot r^2H = sr^2H = r^2H\quad &\sigma_s(3) = 3\\
&s\cdot r^3H = sr^3H = rH\quad &\sigma_s(4) = 2
\end{align*}
Thus \sigma_s = (24). Similarly \sigma_r(1234).
Let G be a group H \leq G, A = \{gh\mid g \in G\}. Let G act on A by left multiplication.
- G acts transitivly on A
- The stabalizer of 1H \in A in G is H
(1) Let aH,bH be elements of A. Let g = ba^{-1}.
g\cdot aH = gaH = ba^{-1}aH = bH
Thus orb(aH) = A so the group action is transitive.
(2) The stabalizer of 1H is
\{g\in G \mid g \cdot 1H = 1H\} = \{g \in G \mid gH = H\} = H
(3) The kernal of the action
\begin{align*}
\ker(A) &=\{g \in G \mid gxH = xH \quad \forall x \in G\}\\
&= \{g \in G \mid x^{-1}gxH = H\quad \forall x \in G\}\\
&= \{g \in G \mid x^{-1}gx \in H \quad \forall x \in G\}\\
&= \{g \in G \mid g \in xHx^{-1} \quad x \in G\}\\
&= \bigcap_{x\in G} xHx^{-1}
\end{align*}
Every group is isomorphic to a subgroup of some symmetric group. If |G| = n then G is isomorphic to a subgroup of S_n.
Let H = 1. Take the permutation representatives
\varphi \colon G \to S_g
By the previous theorem, \ker(\varphi) is contained in H, meaning \ker(\varphi) = e. By the first isomorphism theorem,
g \cong G/\ker(\varphi) \cong \text{im}(\varphi) \leq S_G
If G is a finite group of order n, and p is the smallest prime dividing n, then any subgroup of index p is normal.
Sylow theorem
Recall that Lagranges theorem tells you that the order of a subgroup divides the order of a group. This does not guarentee the existance of a subgroup H_d for every divisor d of a group.
We start with the following Lemma.
If G is a finite abelian group, and p is a prime dividing |G| then G contains an element of order p.
If G is a finite abelian group, and p is a prime dividing |G| then G contains an element of order p.
Use induction on |G|. We know that x \not = e. We have the base case p = |G|. Thus \exists x, x^p = e by Lagranges theorem. Now suppose p < |G|.
If p \mid |x|, then |x| = pn for some n \in \mathbb{Z}
|x^n| = \frac{pn}{\gcd(n,pn)} = \frac{pn}{n} = p
Thus x has order p. If p \not \mid |x|, then let \langle x \rangle = N. Because G is an abelian group, N \trianglelefteq G since N \not = \{e\},
p \mid |G/N|
Thus G/N contains an element of order p. Using the induction hypothesis
|y^p| = \frac{|y|}{p}
Now we get some tricky Definitions
Let G be a group and p be a prime. A group of order p^\alpha for \alpha \geq 0 is called a p-group.
A subgroup of a p-group is called a p-subgroup.
If G is a group of order mp^\alpha where p \nmid m, then a subgroup of order p^\alpha is called a Sylow p-subgroup
Let G be a group. The set of Sylow p-subgroups is denoted
Syl_p(G)
And the number of Sylow p-subgroups
n_p(G)
Sylow had three major theorems in Abstract Algebra relating to these Sylow p-subgroups
Let G be a group of order mp^\alpha where p is a prime not dividing m, then there exists a Sylow p-subgroup.
If P is a Sylow p-subgroup of G and Q is any p-subgroup of G
\exists g \in G \mid Q \leq gPg^{-1}
In particular, any two Sylow p-subgroups are conjugate if Q is a Sylow p-subgroup \exists g such that Q = gPg^{-1}.
The number of Sylow p-subgroups of G is of the form 1 + kp, i.e.
n_p(G) = 1 \mod p
Futhermore, n_p(G) = [G \colon N_g(p)] and n_p(G) \mid m.
We need only prove the first theorem however.
We start our proof using induction on |G|.
For our base case, if |G| = 1, there is nothing to prove. Assume (for the induction hypothesis), Sylow subgroups exist for all groups smaller than |G|.
If p \mid |Z(G)|, by lemma Z(G) has an element of order p and thus a subgroup of order p. Call this subgroup N.
Let \overline{G} = G/N. Note
\frac{|G|}{|N|} = \frac{p^{\alpha}m}{p} = \frac{p^{\alpha - 1}}{m}.
So |\overline{G}| = p^{\alpha - 1}m. By the induction hypothesis, \overline{G} has a subgroup of order p^{\alpha + 1}. Call this subgroup \overline{P}. By the correspondence isomorphism theorem, let P be such that P / N = \overline{P}. Note
|P| = |P/N| |N| = p^{\alpha - 1}p = p^{\alpha}
So P is a Sylow p-subgroup of G.
If p\nmid{}|Z(G)|, let g_1\dots{}g_r be representatives of distinct, non-central congruancy classes of G. Then
|G| = |Z(G)| + \sum\limits^r_{i=1}[G \colon C_G(g_i)]
Note that
\forall i,\,p \mid [G \colon C_G(g_i)] \implies p \mid \sum\limits^r_{i=1}[G \colon C_G(g_i)]
Would imply p \mid Z(G) which we know is false. So therefore there exists some i such that p \nmid [G \colon C_G(g_i)]. For this i, let H = C_g(g_i).
[G \colon C_G(g_i)] = \frac{|G|}{|C_G(g_i)|} = \frac{|G|}{|H|}
So H = p^{\alpha}k for some k such that p \nmid k. Since g_i \not = Z(G), C_G(g_i) \not = G. Therefore k < m. Because |H| < |G| by induction H has a sylow p-subgroup |p| = p^{\alpha}, so P is a Sylow p-subgroup of G.