Math 344 Winter 07
Group Theory Part 3: Quotient Groups
Note: This presentation features 4 uses of the word
“clearly” and one use of the word “clear”. How
many of these are actually clear to you?
Evens and Odds
• We know that (Z, +) is a group.
• We previously observed that this group
could be broken up into two pieces to form a
two element group:
This is a group!
• The elements are subsets of the original
group.
• You add two subsets together like so:
EVENS + ODDS = {n + m n  EVENS & m ODDS}
Flips and Rotations
• We figured out that we could do the same
thing with D8, the group of symmetries of a
square:
This is a group!
• The elements are subsets of D8
• You multiply two subsets together like so:
Flips  Rotations = {ab  aFlips & bRotations}
• In general, given two subsets, A and B, of a
group G we can define their product as:
AB = {ab aA & bB}
We figured out two other ways to make
a 2-element group this way:
• We could break it into:
{I, R2, F, FR2} and {R, R3, FR, FR3}
• Or we could break it into:
{I, R2, FR, FR3} and {R, R3, F, FR2}
Breaking D8 into smaller pieces to make
a group with 4 elements
We only found one way that would work
These groups of subsets are called
quotient groups
• What is needed to be able to construct a
quotient group?
• We found out that:
–
–
–
–
One of the subsets has to be a subgroup
All of the subsets have to be the same size
The subsets can not overlap
All group elements must be used
Cosets
• We figured out how to find all of the subsets
given the subgroup:
– Multiply a group element by the subgroup to get one of
your subsets.
• For example, using the subgroup {I, R2} of D8,
we would:
– Multiply by F (or FR2)to get the subset {F, FR2}
– Multiply by R (or R3) to get the subset {R, R3}
– Multiply by FR (or FR3) to get the subset {FR, FR3}
• These subsets are called the cosets of {I, R2} in
D8.
Coset notation
• Let G be a group and H a subgroup of G.
• Then the set of left cosets of H in G is denoted
by
G/H = {gH  gG}
Where given element g in G the coset gH is
formed by multiplying g by each element of H.
Formally, gH = {gh  hH}
• For practice: Let G = D8, and H = {I, FR}. List
all the cosets of H in G.
Normal Subgroups
• We found out that not all subgroups can be used
to make quotient groups.
– For example, {I, F} cannot be used to make a
quotient group in D8.
• We figured out that we needed gH=Hg for each
element of g in order for H to work.
• Subgroups that satisfy this condition are called
normal subgroups:
Definition: Let G be a group and H a subgroup of
G. Then H is a normal subgroup of G if
gH=Hg gG.
Why is gH = Hg for all gG needed?
Consider the case of H = {I,F} in D8:
When we get to the second spot in row one, we have to
multiply by R on the right of H. This gives us different
stuff than what is in RH = {R, FR3} which was created
by multiplying by R on the left of H!
Formally we can show that we need HggH
for each gG
We need eH gH = gH
For any h in H, ehge is in eH gH.
So we know that hg is in gH. Thus HggH
In the finite case, we can say that Hg and gH are clearly
the same size so they must in fact be the same sets:
Hg=gH.So H must be normal.
Question you don’t have to answer: Is it necessary to
also have gHHg in the infinite case or is HggH
good enough? Yes! Proved in just a few slides!
An easier way to multiply cosets:
Definition: Let G be a group and H a subgroup of G. If a, b
G, then:
aH bH = abH.
Does this match our original way of thinking about
multiplying cosets?
First recall that gH = {gh: h H}
Then, by our original definition of multiplication of
subsets,
aH bH = {(ah)(bh )  h, h H}
But by our new definition,
aH bH = abH = {(ab)h  h H}.
Show that if H is normal,
{(ah)(bh )  h, h H} = abH.
Showing our old definition is the same as the
new one (if the subgroup is normal!)
Let G be a group and H a normal subgroup of G. Will
show that
{(ah)(bh )  h, h H} = abH.
First let x  abH. Then x = abh1 for some h1H.
If we let h2 = e, we have x = aebh1 = (ah2)(bh1). This
means that x  {(ah)(bh )  h, h H}
Second, we let x  {(ah)(bh )  h, h H}, we have x =
(ah1)(bh2) for some h1, h2 H. Since H is normal, bH =
Hb. Clearly h1b Hb. Then we also have h1b bH. This
means that h1b = bh3 for some h3 H. Thus x =a(h1b)h2
=a(bh3)h2 =(ab)(h3h2 ). Since H is a subgroup and hence
closed, h3h2 H. Therefore xabH.
This proves that {(ah)(bh )  h, h H} = abH.
So in the case of a normal subgroup, our old way of
thinking about multiplying cosets matches our new
definition. Yeah!
An important detail
Lemma X: Let G be a group and H a subgroup of G.Let a, b
G. Then baH  aH = bH (and bHa  Ha = Hb.)
Proof. “” Suppose that baH. Then b = ah1 for some h1H.
Now, let xaH. Then x = ah2 for some h2H. But since b =
ah1, a = bh1-1. Substituting, we have x = b(h1-1 h2). Since H
is a subgroup, the stuff in parentheses is just an element of
H, so xbH. We have shown aH  bH.
Now, let xbH. Then x = bh2 for some h2H. Substituting we
get x = a (h1h2). Again since H is a subgroup, the stuff in
parentheses is just an element of H, so xbH. We have
shown aH  bH.
Therefore, aH = bH.
“” Suppose that aH = bH. Clearly aaH since a = ae where
e is the identity of G. But then since aH = bH, we must also
have abH
Thus baH iff aH = bH (The proof that bHa iff Ha = Hb is
virtually identical.)
What does this lemma actually mean?
Lemma X: Let G be a group and H a subgroup of G.Let a,b G.
Then baH  aH = bH and bHa  Ha = Hb.
This means that if two cosets have anything in common, they
are exactly the same. And this works for both left and right
cosets.
This captures our informal idea that the cosets do not
overlap.
Note that this is true even if the subgroup is not normal
(check out the proof, no use of normal!)
Do we need H to be normal to construct
a quotient group?
We can prove easily that we need to have HggH
for each gG.
Proof: Let gG. Let hH. Since eH =hH, we must have
eHgH = hHgH. Thus we must have gH = hgH. But
then by our lemma we must have hg gH. Thus we
must have HggH for each gG.
In the finite case this means we must have Hg=gH for
each gG. So H must be normal!
Question you don’t have to answer (same one as before)
in the infinite case, do we need to also have gHHg
for each gG? The answer is YES! See next slide…
Why it actually has to be normal!
You only need the condition Hg  gH to make a
quotient group (as we shall see), but this condition
implies that gH  Hg as well! So if G/H makes a
group, then H must be normal.
Here is the proof: Suppose G is a group and H is a
subgroup of G. Suppose that Hg  gH for all gG.
Let gG. Then we know in particular that Hg-1  g-1H.
Let x  gH (Just need to show x  Hg). Then x=gh for
some h  H.
Note that hg-1  Hg-1. Thus we have hg-1  g-1H. This
means that hg-1 = g-1h1 for some h1 H. But then
ghg-1 = gg-1h1 = h1. (Almost there!) Then ghg-1g =
h1g. Therefore x = gh = h1g. So we have x  Hg.
Thus gH  Hg for all gG.
We have shown that if Hg  gH for all gG. then in
fact Hg = gH for all gG.
Woo Hoo! (or QED)
Theorem: A subgroup can always be used to
construct a quotient group if it is normal!
Theorem: Let G be a group and H a subgroup. Then G/H is a group
under the operation defined above if and only if H is normal.
Proof: Suppose that H is a normal subgroup of G.
First we need to show that aHbH = abH a, bG. gives an operation
defined on G/H.
Everywhere defined: This is clear, given two cosets, the formula
gives a coset!
Well-Defined: Need to show that if aH = cH and bH = dH, then
aHbH=cHdH. (This is one of the HW problems).
By definition, aHbH=abH and cHdH=cdH. So we need to show that
abH =cdH. By our previous lemma, we just need to show that cd
abH
Since aH = cH then c aH. Similarly, we have d bH. Then, c = ah1
and d =b h2 for some h1, h2 H. Then cd = ah1bh2 = a(h1b)h2
Clearly h1b Hb. But since H is normal, bH= Hb. Thus h1b bH.
So, h1b = bh3 for some h3 H. This gives cd = a(bh3)h2 = (ab)(h3h2).
Thus cd abH (since h3h2 bH.). So our operation is well
defined.
Theorem: A subgroup can always be used to
construct a quotient group if it is normal!
Theorem: Let G be a group and H a subgroup. Then G/H is a group under
the operation defined above if and only if H is normal.
Proof: “ cont.” Now that we know we have defined an operation on the set
of left cosets, we need to check that it gives us a group. This is the easy
part! Let gG and e be the identity of G.
Identity: From the definition of our operation and the existence/definition of
the identity of G, we have,
eH gh = egH = gH = geH = gH eH. So eH is the identity of G/H.
Inverses: From the definition of our operation and the existence/definition of
inverses in G, we have,
gH g-1H = g-1H gH = eH. So each coset has an inverse in G/H.
Associative: Let a, b, c G. From the definition of our operation and the
associativity of G we have,
aH (bH cH) = aH (bc)H = a(bc)H
=(ab)cH = (ab)H cH = (aH bH ) cH
So if H is a normal subgroup, G/H is a group!
Note:If we had stuck with our old definition of coset multiplication, we
would have had to use normality to prove this part as well, and the proof
would have been more complicated!