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Math 542
Day 8 Follow-up
All About Quotient Thingies
Part 1: A handy little lemma about cosets
Lemma: Given a group G and a coset H, for any elements a, b in G, the cosets aH and bH
are equal if and only if b-1a is an element of H. In math shorthand: aH = bH ⇔b-1a∈H.
Proof: (⇒) Suppose that aH = bH. Note that a∈aH, so now we know that a∈bH. Then
there exist h∈H such that a = bh. Solving for h we see that h is b-1a. Thus b-1a∈H.
(⇐) Suppose that b-1a∈H. First, let x ∈aH. This means we have h1 ∈H such that x = ah1.
Now since b-1a∈H we have h2 ∈H such that b-1a =h2. Solving for a, we get a = bh2.
Substituting we get x = bh2h1. Thus x ∈bH. Second, let x ∈bH. If we do the same stuff
but this time using the fact that a-1b∈H (it is the inverse of b-1a and H is a subgroup) we
get b-1a. Thus aH = bH.
What is this lemma good for? Well without it, whenever we needed to prove two cosets
(aH and bH) were equal, we would need to do a set equality proof (like the one we just
did), now we just need to verify that a single element is in the subgroup (b-1a).
What does this lemma look like when the cosets are additive? Well the cosets now look
like a + H and b + H and b-1 now looks like -b. Instead of multiplying b-1 and a, we will
add -b and a to get a-b. So it looks like: a + H = b + H ⇔ a-b ∈H.
Part 2: Showing that normality is exactly what we need for quotient groups to work
Intro: The elements of a quotient group are the cosets of a subgroup (turns out we need it
to be normal). So (using multiplicative notation) the elements of the quotient group G/N
are things that look like gN where g is an element of G. The operation for a quotient
group is given by g1N g2N = g1 g2N.
Where it gets interesting: Cosets can have multiple names. As our handy little lemma
tells us, any time b-1a is an element of N, we have aN = bN. This could be a big problem.
Suppose I wanted to multiply the coset aN by the coset cN. I would get acN. But if you
multiplied the same two cosets but used the name bN instead of aN, you would get bcN.
If the cosets acN and bcN are different we are in BIG TROUBLE because our operation
needs to give the same answer if we are operating on the same things! What we need to
do is figure out what condition is necessary and sufficient to ensure that our coset
multiplication is well-defined.
(Necessary) Let’s proceed by finding a condition that is needed and then proving that it is
good enough. So suppose that coset multiplication is well-defined. Now we know that for
any n ∈ N, the coset nN is the same as the coset eN where e is the identity of G. This
means that for any g ∈ G, we need to get the same answer for nN gN and eN gN. Using
our coset multiplication definition, this means that ngN = gN. But then our handy little
lemma tells us that g-1ng ∈N. So this gives us a necessary condition. Let’s define any
subgroup that satisfies this property to be a normal subgroup.
Definition: Let G be a group and N a subgroup of G. Then N is a normal subgroup of G if
for every g ∈G and n ∈N, g-1ng ∈N.
Note that normality is not a property of N as a group itself, but is a property it has in
relation to G.
(Sufficient) We have shown that if our coset multiplication is well defined we must use a
normal subgroup. It would be really cool it turned out that normality also guaranteed that
our coset multiplication was well defined. So let’s try to prove that.
Let N be a normal subgroup of G. Now suppose that we have elements a, b, c, d in G
such that aN = cN and bN = dN. We need to show that aN bN = cN dN. But using our
definition of coset multiplication, this is equivalent to showing that abN = cdN. And
using the handy little coset lemma, this is equivalent to showing that (cd)-1ab is an
element of N. Let’s play with this element and see if we can prove it is an element of N:
(cd)-1ab = d-1c-1ab = d-1 (c-1a) b.
Now since aN = cN we know that c-1a is an element of N, call it n1. So now we have:
(cd)-1ab = d-1 n1 b = d-1 (n1 b).
Here is where we will use normality. Since N is normal, we know that b-1n1b is an
element of N, call it n2. Multiplying both sides of the equation b-1n1b = n2 by b (on the
left), we get n1b =bn2. (This is why normality is sometimes thought of as quasicommutivity, we can commute elements of the form gn as long as we are aware that
element from N could change.)
Now we substitute to get (cd)-1ab = d-1(bn2) = (d-1b)n2. And since bN = dN, our handy
little coset lemma tells us that d-1b is an element of N. Thus we have that (cd)-1ab is an
element of N, which means that abN = cdN and so our coset multiplication is well
defined!
So there you have it, normality is a necessary and sufficient for coset multiplication to be
well defined. It is now very easy to verify that the set of cosets of a normal subgroup
form a group under coset multiplication (give it a go!).
Part 3: Now for Rings
First note that for any commutative (abelian) group, every subgroup is normal.
Proof: If N a subgroup of an abelian group G, then for any n in N and g in G, ng = gn,
and so g1ng = n and so is an element of N.
Remember that a ring is an abelian group under addition. This means that any additive
subgroup of a ring R is a normal subgroup. So addition of additive cosets is always going
to be well defined for a ring. (Note that here we are just forming cosets out of any old
additive subgroup – at this point it is not clear that we aught to be using a subring).
So if we can figure out a way for multiplication of additive cosets to be well defined we
will be well on our way to forming a quotient ring.
But first, why not use multiplicative cosets? These have many problems starting with the
fact that they won’t partition the ring into equal sized non-overlapping pieces. For
example, take any subgroup S (or if you wish a subring) of a ring R. Note that the
multiplicative coset 0S = {0} while the multiplicative coset 1S = S. So these two cosets
are of different size if S is non-trivial and they overlap as well because 0 is in S.
Of course the other problem is that if we used multiplicative cosets, we would not be
taking advantage of the fact (noted above) that additive cosets automatically have one
well defined operation (addition).
So what about multiplication of additive cosets?
Well we would surely define (a + S) (b + S) to be ab + S. Now let’s try to find a
condition that is necessary and sufficient to ensure that this multiplication is well defined.
Part 4: Ideals of a ring and why they are so great for making quotient rings.
(Necessary Condition). Let R be a ring. Suppose we form additive cosets of a subgroup S
of R and suppose that coset multiplication is well defined. Then since for any element s in
S, the cosets 0 + S and s + S are equal, we know that (0 + S) (r + S) = (s + S) (r + S) for
any element r in R. But this means that 0 + S = sr + S. Using our handy little coset
lemma, we know that sr must be an element of S.
Of course we also must have (r + S) (0 + S) = (r + S) (s + S), we we also know that rs
must be an element of S.
So if multiplication of additive cosets is well defined we must have that for any s in S and
r in R, both sr and rs are elements of S. Let’s give this condition a name!
Definition: An additive subgroup S of a ring R is an ideal if for each s ∈S and r ∈R, we
have sr ∈S and rs ∈S. (or equivalently: for any r in R, rS ⊆ S and Sr ⊆ S.)
Before, proving that multiplication of additive cosets of an ideal is always well defined,
let’s address the question of whether we are using subgroups or subrings to make quotient
rings.
Claim: An ideal S of a ring R is a subring of R.
Proof: An ideal is defined to be a subgroup, so we need only establish that S is closed
under multiplication. But ideals actually satisfy a MUCH STRONGER condition. Since
every product ab in R is an element S as long as at least one of a and b is in S. So clearly
if both a and b are in S, then ab is in S.
And ideal is kind of like a black hole. Every product that involves an element of the ideal
is sucked into the ideal.
So the assumption that S is an ideal of R also means that S is a subring of R.
(Sufficient). Now let’s prove that multiplication of additive cosets of an ideal S of a ring
R is well defined.
Proof: Let S be an ideal of a ring R. Suppose that a + S = c + S and b + S = d + S. We
need to prove that (a + S) (b + S) = (c + S) (d + S).
So what I need to do is show that ab + S = cd + S which I can do by showing that cd-ab
is in S (handy little coset lemma).
First I will employ the ubiquitous technique of adding zero cleverly:
cd – ab = cd – ad + ad – ab = (c-a)d + a(d-b).
Now since a + S = c + S, we know that c – a is in S, say c – a = s1. Similarly d - b is in
S, say d – b = s2. Thus cd – ab = s1d + a s2. Now since S is an ideal, both s1d and a s2 are
in S. And since S is closed under addition, the sum s1d + a s2 is in S. So we have cd-ab is
in S which means that ab + S = cd + S and so multiplication of additive cosets of S is well
defined!
There it is, all the tough stuff about quotient groups and quotient rings. The rest should all
be fun and games!