Subsection 1.3.1 Morphisms
Having studied groups, rings, vector spaces and similar objects, you have considered the notion of a homomorphism, a structure-preserving map. While the notions of a linear map, group or ring homomorphism differ in the details, they all are defined to preserve whatever structure the algebraic object has. For example,
A
linear map \(T:V\to W\) between two vector spaces over the same field
\(F\) requires that for all
\(v,v^\prime
\in V\) and
\(\lambda \in F,\)
\begin{equation*}
T(v+ v^\prime) = T(v) + T(v^\prime) \text{ and }
T(\lambda v) = \lambda T(v),
\end{equation*}
transporting the structure of vector addition and scalar multiplication on
\(V\) to the corresponding ones on
\(W.\)
A
group homomorphism \(\varphi: G \to H\) between groups
\(G,H\) requires that for all
\(g, g^\prime
\in G\text{,}\)
\begin{equation*}
\varphi(g*g^\prime) = \varphi(g)*
\varphi(g^\prime)
\end{equation*}
transporting the binary operation on
\(G\) to the one on
\(H.\)
A
ring homomorphism \(\varphi:R\to S\) between rings
\(R,S\) requires that for all
\(r,
r^\prime\in R\)
\begin{equation*}
\varphi(r+r') = \varphi(r) + \varphi(r') \text{ and }
\varphi(r*r^\prime) = \varphi(r)*\varphi(r^\prime),
\end{equation*}
transporting the additive abelian group structure and multiplicative structure on
\(R\) to the corresponding structures on
\(S.\)
The subject of category theory generalizes the notion of a homomorphism to an extreme providing a definition of a morphism \(\varphi:X\to Y\) between objects \(X\) and \(Y,\) where the objects need not be restricted to algebraic objects, but could include things like topological or analytic spaces.
In this very general context, one defines the notion of an isomorphism, as a morphism \(\varphi:X\to Y\) for which there is a morphism \(\psi:Y\to X\text{,}\) so that \(\varphi\circ \psi = id_Y\) and \(\psi\circ\varphi
=id_X.\) In particular, this means that \(\varphi\) is a bijective morphism.
The bottom line is that we catch a break in dealing with algebraic homomorphisms, and need only verify that our chosen homomorphism is bijective.
Subsection 1.3.2 Quotients
Given an algebraic object \(X\) and a sub-object \(Y\) (e.g., group and subgroup, ring and subring, vector space and subspace), we can ask whether the quotient \(X/Y\) is again an algebraic object of the same type.
The notion of a quotient is certainly deeply connected with that of homomorphisms, the most universal theorem being the first isomorphism theorem, which for a homomorphism \(\varphi: X\to Y\) always has the form
\begin{equation*}
X/\ker \varphi \cong \Im \varphi
\end{equation*}
In general the kernel,
\(\ker \varphi\text{,}\) always has the right properties to guarantee that the quotient
\(X/\ker\varphi\) is an algebraic object of the same type. While it is always true if
\(W\) is a subspace of a vector space
\(V,\) that
\(V/W\) is again a vector space, quotients do not always have the desired properties. Recall for groups, one needs a subgroup
\(H\) to be a
normal subgroup of a group
\(G\) for the set of cosets
\(G/H\) to be a group, and we need
\(S\) to be an
ideal of a ring
\(R\) for the quotient
\(R/S\) to be a ring. Indeed these definitions arise and are natural precisely because they describe the conditions under which a quotient will have the appropriate structure.
Subsection 1.3.3 Cosets, partitions, and equivalence relations
Meeting the notion of a coset for the first time often seems to cause a bit of confusion, some of which arises from notation (additive or multiplicative), and some of which arises from failing to understand the underlying equivalence relation. Let’s try to understand all the issues.
There are couple of considerations. The first is that vector spaces and rings have underlying group structures so that when we form cosets with them, they are first and foremost group cosets. The set of cosets may enjoy additional structure depending on the context, but what sets the notation — and the equivalence relation — is the group structure.
Second in most (though not all) introductions to abstract algebra, groups enter the picture first, and being a set with a single binary operation, it is very often written multiplicatively. This works well in our experience since in most settings (think rings) addition is commutative, but multiplication may not be (e.g., matrices). So introducing notation for a group that may or may not be abelian, multiplication is generally the more intuitive choice. So for a group \(G\text{,}\) subgroup \(H\text{,}\) and element \(x\in
G\text{,}\) we write \(xH\) or \(Hx\) depending on whether we are talking about left cosets or right cosets.
The complication arises when we have a ring \(R\) and an ideal \(I\) and want to discuss the structure of the set of cosets \(R/I.\) Now we need both operations of multiplication and addition, so choosing a multiplicative notation for the groups structure would lead to a notational nightmare. Moreover, the additive groups structure is now abelian, so it is a bit more natural to use additive one for cosets, so we write \(x+I\) (or \(I+x\) though they turn out to be the same since addition is commutative).
Now that we have some sense of why we choose the notation, let’s understand the equivalence relation and the associated partition. Since all the cosets start out with an underlying group, we use group notation (both multiplicative and additive) to describe the cosets and equivalence relation.
Let \(G\) be a group, \(H\) a subgroup, and \(x\in
G.\) We define left cosets:
\begin{align*}
xH \amp = \{y \in G \mid y = xh \text{ for some } h\in H\}
= \{xh \mid h \in H\}\\
x+H \amp = \{y \in G \mid y = x+h \text{ for some } h\in H\}
= \{x+h \mid h \in H\}
\end{align*}
We denote the set of left cosets by
\begin{equation*}
G/H = \{ xH \mid x\in G\} \text{ or } \{ x+H\mid x\in
G\}
\end{equation*}
depending upon whether we are using multiplicative or additive notation.
Now it is obvious that each coset is a subset of \(G,\) and what you prove is that the set of left cosets forms a partition of \(G.\) Since the identity of the group, \(e\text{,}\) is contained in every subgroup \(H,\) every element \(x\in G\) is an element of the coset \(xH\) since \(x =
xe \in xH.\) It follows that
\begin{equation*}
G = \bigcup_{x\in G} xH,
\end{equation*}
the first property of a partition.
The critical condition is that cosets are either disjoint or identical which makes their collection a partition.
Proposition 1.3.2.
The (left) cosets of \(H\) in \(G\) are either disjoint or identical.
Proof.
Suppose we have two cosets \(xH\) and \(yH.\) If they are disjoint that is fine, but if they intersect, we must show that are equal. Let’s do this out in full detail so we better understand the rule we shall write down for determining whether or not two cosets are equal.
Let \(z \in xH\cap yH.\) To show that \(xH = yH\) (recall they are sets), we must show \(xH \subseteq yH\) and \(yH \subseteq xH\text{.}\) The argument we give will be symmetric, so we need only show one containment, say \(xH
\subseteq yH.\)
The condition \(z \in xH\cap yH\) says that we may write
\begin{equation*}
z = xh_1 = yh_2 \text{ for some } h_1, h_2 \in H,
\end{equation*}
so \(x = y h_2 h_1^{-1} \in yH,\) and so of course every element \(xh \in xH\) is equal to \(xh = y h_2 h_1^{-1}h \in yH\) which gives the desired conclusion \(xH \subseteq yH.\)
What we must also understand is that while
\begin{equation*}
G = \bigcup_{x\in G}
xH
\end{equation*}
there is a great deal of redundancy in the (multi)set \(\{xH\mid x\in G\}\text{.}\) An example should help.
Example 1.3.3.
Let \(G = S_3\text{,}\) the symmetric group on 3 letters, i.e., the permutations of the set \(\{1,2,3\}.\) Let \(H=\{
e,(1\ 2)\}\) be the cyclic subgroup generated by the transposition \((1\ 2)\text{.}\)
Solution.
Then the set of left cosets is
\begin{equation*}
G/H=\{H, (1\ 3)H, (2\ 3)H\}=\{H,(1\ 2\ 3)H,(1\ 3\ 2)H\}
\end{equation*}
where
\begin{equation*}
H = (1\ 2)H,\quad (1\ 3)H = (1\ 2\ 3)H, \text{ and } (2\
3)H = (1\ 3\ 2)H.
\end{equation*}
Note that the associated partition of \(G\) is
\begin{equation*}
\mathcal P_L = \{ P_1=\{1, (1\ 2)\}, P_2=\{(1\ 3),(1\ 2\
3)\},
P_3=\{(2\ 3), (1\ 3\ 2)\},
\end{equation*}
where the \(P_i\) are simply the elements in the respective cosets.
For the record we record the set of right cosets and their associated partition which differs from the one from left cosets (since \(H\) is not a normal subgroup of \(G\)).
\begin{equation*}
H\backslash G = \{H, H(1\ 3), H(2\ 3)\} = \{H, H(1\ 3\ 2), H(1\ 2\
3)\}
\end{equation*}
The associated partition is of \(G\) is
\begin{equation*}
\mathcal P_R = \{Q_1=\{1, (1\ 2)\}, Q_2 = \{(1\ 3), (1\
3\ 2)\}, Q_3=\{(2\ 3), (1\ 2\ 3)\}.
\end{equation*}
To distinguish when two cosets are equal or the same, we have the following criterion.
Proposition 1.3.4.
Let \(G\) be a group and \(H\) a subgroup. Then two (left) cosets \(xH\) and \(yH\) are equal if and only if and of the following equivalent conditions hold:
\(\displaystyle y^{-1}x \in H\)
\(x=yh\) for some \(h\in H\)
\(\displaystyle x^{-1}y\in H\)
\(y = xh\) for some \(h\in H\)
Proof.
It is quite straightforward to show that these four conditions are all equivalent, but the important part is why they are equivalent to \(xH=yH.\)
The key to that is to remember
Proposition 1.3.2, that cosets are disjoint or equal, so to show that
\(xH = yH\text{,}\) one needs only show that
\(xH \cap yH \ne \emptyset\text{.}\) So it is enough to show that
\(x \in yH\) or
\(y \in
xH\) which lead naturally to the conditions in the proposition.
Subsection 1.3.4 Introducing an algebraic structure on the set of cosets.
Whether we are talking about the quotients of groups, rings, or vector spaces, there is always an underlying group \(G\) and a subgroup \(H.\) For quotient groups this is obvious; for rings, the group is the additive group of the ring, and for vector spaces, the group is the additive group of the vector space.
To make a set of cosets into a group, we need \(H\) to be a normal subgroup of \(G\text{.}\) Recall that
Proposition 1.3.6.
Let \(G\) be a group and \(H\) a subgroup of \(G.\) The following conditions are equivalent and define what it means for \(H\) normal subgroup of \(G\text{.}\)
\(gHg^{-1} = H\) for all \(g \in G\)
\(gH =Hg\) for all \(g \in G\)
\(gHg^{-1} \subseteq H\) for all \(g \in G\)
Proof.
The equivalence of the first two is easy to check, and of course the first implies the third, so we are left to show that \(gHg^{-1} \subseteq H\) for all \(g \in G\) implies that \(gHg^{-1} = H\) for all \(g \in G\text{.}\)
Fix a \(g\in G\) for which \(gHg^{-1} \subseteq
H\text{;}\) we must show the reverse inclusion \(H \subseteq
gHg^{-1} \text{.}\) Since \(xHx^{-1} \subseteq H\) for all \(x \in G\text{,}\) choose \(x=g^{-1}.\) Then
\begin{equation*}
g^{-1}H g \subseteq H \text{ which implies }
H \subseteq gHg^{-1}\text{,}
\end{equation*}
which is the desired inclusion.
Checkpoint 1.3.7. Why is normality the correct notion to make a set of cosets into a group?
Let’s make sense of the question. Our set of left cosets is
\begin{equation*}
G/H = \{xH \mid x\in G\}\text{.}
\end{equation*}
Our job (still working multiplicatively) is to define the product of two cosets in a well-defined manner:
\begin{equation*}
xH \cdot yH = zH
\end{equation*}
for some \(z\in G.\) Now everybody knows that they want the answer to be
\begin{equation*}
xH \cdot yH = xyH,
\end{equation*}
but it is not clear that always makes sense.
Let’s go back to our example with \(G=S_3\) and \(H = \{e, (1\ 2)\}.\) We observed that the cosets paired up as
\begin{equation*}
H = eH=(1\ 2)H,\quad (1\ 3)H = (1\ 2\ 3)H, \text{ and } (2\
3)H = (1\ 3\ 2)H.
\end{equation*}
So to be well-defined, whatever definition we come up with cannot depend on how we name the sets, so in our case it must be true that (for example)
\begin{equation*}
(1\ 2)H \cdot (1\ 3)H= e H \cdot (1\ 2\ 3)H\text{,}
\end{equation*}
but if we used our desired rule we would get
\begin{equation*}
(1\ 2)H \cdot (1\ 3)H = (1\ 3\ 2)H \text{ while }
e H \cdot (1\ 2\ 3)H = (1\ 2\ 3)H
\end{equation*}
which are not equal.
The exercise is to see how the definition of normality arises naturally in trying to rectify this ambiguity.
Similarly, given a ring \(R\) and a subring \(S\text{,}\) we know that the set of cosets
\begin{equation*}
R/S = \{r + S \mid r\in R\}
\end{equation*}
is an abelian group under addition, but what about multiplication? Can we make the set of cosets into a ring?
Checkpoint 1.3.9. Why is the property of being an ideal the correct property to make \(R/S\) into a ring?
For any subring \(S\) of \(R\) we already have the quotient group \(R/S = \{ r+S \mid r\in R\}.\) It is natural to define multiplication as
\begin{equation*}
(r+S)\cdot (r^\prime+S) = rr^\prime + S,
\end{equation*}
but we need to check that things are well-defined.
To do so we need to ensure that for \(s,t \in S\text{,}\)
\begin{equation*}
(r+s+S)\cdot (r^\prime + t + S) = rr^\prime + S,
\end{equation*}
but our would-be definition tells us we would need
\begin{equation*}
rr^\prime + rt + sr^\prime + st + S = rr^\prime +S.
\end{equation*}
Since \(S\) is subring we know that \(st \in S,\) but we also need that \(rt + sr^\prime \in S\) for any \(s,t
\in S.\) Thus we arrive at the condition that \(S\) be a two-sided ideal of \(R\text{,}\) namely that in addition to being a subring of \(R\text{,}\) we must have
\begin{equation*}
R\cdot S \subseteq S \text{ and } S\cdot R \subseteq S.
\end{equation*}
