Factor Groups
Table of Contents
1. Cosets and Groups
Let \( (G , \ +) \) be a group and consider some subgroup \( H \le G \), we start by defining:
\[ g + H = \{ g + h : \forall h \in H \} \text{ for some } g \in G \]
We say \( g + H \) is a left coset of \( H \) in \( G \), right cosets are defined similarly.
Lemma 1: The function \( f: H \to G, \ h \mapsto g + h \) is injective
This follows immediately from the cancellation property of groups. From this it is apparent that (in both the finite and infinite cases):
\[ \left|H\right| = |g + H| \ \forall g \in G \]
1.1. Disjointedness
We now prove an extremely important property of cosets. If \( g_1, \ g_2 \in G \) then exactly one of the following is true:
\begin{align*} 1) \ &g_1 + H = g_2 + H \\ 2) \ &g_1 + H \cap g_2 + H = \emptyset \end{align*}Ie two cosets are either equal or disjoint. From this we can deduce that the cosets of \( H \) partition \( G \), and thus arrive at Lagrange's Theorem in the finite case.
1.1.1. Proof
Now suppose that \( g_1 + H \) and \( g_2 + H \) share some common element \( k \), and further suppose for a contradiction that there is some element \( q \in G \ s.t. \ q \in g_1 + H, \ q \not\in g_2 + H \) ie \( q = g_1 + h_3 \) for some \( h_3 \in H \). Then:
\begin{align*} \exists h_1, h_2 \in H \ s.t. \ k &= g_1 + h_1 \\ k &= g_2 + h_2 \\ \end{align*}Equating these gives:
\begin{align*} &\implies g_1 = g_2 + h_2 + (-h_1) & \\ &\implies q = g_2 + h_2 + (-h_1) + h_3 & \\ &\implies q \in g_2 + H &\text{ since } h_2 + (-h_1) + h_3 \in H \end{align*}Which is a contradiction, hence there exists no such element \( q \). From this it is apparent that if the two cosets share an element, they must be subsets of one another. This completes the proof.
1.2. Cosets as Groups
Suppose we want to create a group whose elements are the cosets of \( H \), and which operate under the "same" binary operation as \( G \), ie
\[ (a + H) + (b + H) = (a + b) + H \]
The problem here is that one coset \( g + H \) may have many representations, so the operator may not be well defined,
1.2.1. example \( S_3 \)
Let \( G = S_3 \), \( H = <(1 \ 2)> = \{ (1 \ 2), \ I_3 \} \) then our cosets are as follows:
\[ \{(1 \ 2), \ I_3 \}, \ \{ (1 \ 3), \ (1 \ 2 \ 3) \}, \ \{ (2 \ 3), \ (3 \ 2 \ 1)\} \]
If we follow our definition of the binary operation, then:
\[ \left( (1 \ 2 \ 3) * H \right) * \left( (3 \ 2 \ 1) * H \right) = I_3 * H \]
But
\[ \left( (1 \ 3) * H \right) * \left( (3 \ 2 \ 1) * H \right) = (2 \ 3) * H \]
Hence in this case the result of the operation on the cosets depends on the element we use to generate it.
We now find conditions in which multiplication of cosets is independent of the element we use to generate the coset.
1.2.2. Independent of Choice
Let \( y \in G \), \( h_1 \in H \), and take any \( x, b \in G \) s.t.
Suppose \( a, \ b, \ x, \ y \in G \) are such that the following is true:
\begin{align*} a + H &= b + H \\ x + H &= y + H \end{align*}We want \( (a + H) + (x + H) = (b + H) + (y + H) \) (ie \( (a + x) + H = (b + y) + H \) by our definition of the group operation), so that our choice of generator does not affect the product. We have:
\begin{align*} a &= b + h_1 \\ x &= y + h_2 \end{align*}Recall \( (a + x) + H = (b + y) + H \iff (a + x) \in (b + y) + H \) now rearranging:
\begin{align*} (a + x) &= b + h_1 + y + h_2 \\ &= b + y + (-y) + h_1 + y + h_2 \\ \end{align*}Hence \( (-y) + h_1 + y \in H \implies (a + x) \in (b + y) + H \). Thus we have if \( H \) is closed under conjugation ( a similar, but weaker condition than commutativity) independence of choice holds.
We show by contrapositive the converse also holds, namely if independence of choice holds, then \( H \) is normal.
Suppose \( H \) is not closed under conjugation, ie there exists \( y \in G, \ h \in H \) s.t. \( (-y) + h + y \not \in H \).
\begin{align*} 0 + H &= (0 + h_1) + H \\ y + H &= y + H \end{align*}Suppose also that \( (y + 0) + H = (h_1 + y) + H \).
\[ \implies y \in (h_1 + y) + H \\ \implies \exists h_2 \ s.t. \ y + h_2 = h_1 + y \\ \implies h_2 = (-y) + h1 + y \]
A contradiction. Thus independence of choice holds iff \( H \) is normal in \( G \).
2. Normal Subgroups
We can now claim if \( H \) is normal, the cosets (we will see that whether they are left or right cosets is not important) form a group under the "same" binary operation of \( H \).
Invertability and existence of an identity all follow by virtue of \( G \) (to be more specific, using the coset definition: \( a + H = \{ ah : h \in H \} \) readily proves these claims).
2.0.1. Proposition
\( N\triangleleft G \) iff \( g + N = N + g \ \ \ \forall g \in G \). Ie the right and left cosets of \( N \) in \( G \) are identical.
2.0.2. Proof
Suppose \( k \in g + N \) then:
\begin{align*} \exists h \in N \ s.t. \ k = gh \\ \implies k = (ghg^{-1})g \\ \end{align*}But \( g + h + (-g) \in N \) hence \( k \in N + g \). And similarly \( k \in N + g \implies k \in g + N \).