Math 344 Winter 07Group 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 aFlips & bRotations}

• In general, given two subsets, A and B, of a group G we can define their product as:

AB = {ab aA & bB}

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 gG}

Where given element g in G the coset gH is formed by multiplying g by each element of H.

Formally, gH = {gh hH}

• 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 gG.

Why is gH = Hg for all gG 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 HggH for each gG

We need eH gH = gHFor any h in H, ehge is in eH gH. So we know that hg is in gH. Thus HggH

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 gHHg in the infinite case or is HggH 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 h1H.

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 xabH.

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 baH aH = bH (and bHa Ha = Hb.)

Proof. “” Suppose that baH. Then b = ah1 for some h1H.

Now, let xaH. Then x = ah2 for some h2H. 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 xbH. We have shown aH bH.

Now, let xbH. Then x = bh2 for some h2H. Substituting we get x = a (h1h2). Again since H is a subgroup, the stuff in parentheses is just an element of H, so xbH. We have shown aH bH.

Therefore, aH = bH. “” Suppose that aH = bH. Clearly aaH since a = ae where

e is the identity of G. But then since aH = bH, we must also have abH

Thus baH iff aH = bH (The proof that bHa 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 baH aH = bH and bHa 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 HggH for each gG.

Proof: Let gG. Let hH. 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 HggH for each gG.

In the finite case this means we must have Hg=gH for each gG. 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 gHHg for each gG? 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 gG.

Let gG. 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 gG. We have shown that if Hg gH for all gG. then in fact

Hg = gH for all gG.

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, bG. 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 gG 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 ) cHSo 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!