Quotient Groups
Lesson 2
• Task 2
We made some conjectures as to what our identity subset needed to have in order for
our subsets to form a group.
1. Does the identity element of a quotient group need to contain the identity of the
original group? Why? (U3L2T3-Handout2-Identity.pdf)
2. Does the identity element of a quotient group need to be a closed subset? Why?
(U3L2T3-Handout1-Closed.pdf)
3. Does the identity element of a quotient group need to contain the inverse of each
of its elements? Why? (U3L2T3-Handout3-Inverses.pdf)
• Task 4
Suppose you know which subgroup you want to use for the identity of your quotient
group. How do you figure out what the other subsets need to be? Find out how the
rest of the group will be partitioned given the following subgroups:
{I, F R}
{I, R2 }
• Task 5
Definition. Let G be a group, H a subgroup, and g an element of G.
Left coset: gH = {gh | h ∈ H}
Right coset: Hg = {hg | h ∈ H}
• Task 6
Given a subgroup and its cosets, will we always be able to construct a quotient group?
Why does the subgroup {I, R2 } and its cosets form a quotient group but the subgroup
{I, F R} and its cosets fail to form a quotient group? (U3L2T6-Handout4-WhyQG.pdf)
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Lesson 3
• Task 1
We are trying to figure out what conditions are needed to make a quotient group. Let’s
summarize what we have found so far:
1. When we partition the group we want to use all of the group elements. (It is
possible to make a quotient group using only part of the group if the part you
break up is a subgroup).
2. The subsets that are the elements of our quotient group all have to be the same
size. (Try it with different sized subsets at home for fun - enjoy the chaos).
3. The subsets that are the elements of our quotient group can’t overlap. (Try it
with overlapping subsets at home for fun - enjoy the chaos.)
4. The identity subset of a quotient group must be a subgroup. (Why? We proved
it last time).
5. We figured out that the other subsets must be cosets of the subgroup. (We
obtained these cosets by multiplying an element of the group by each element of
the subgroup).
6. We figured out that the cosets of a subgroup will not form a quotient group unless
the left and right cosets are the same.
• Task 2
Definition. Let G be a group and H a subgroup. Then H is a normal subgroup of G
if gH = Hg for all g ∈ G. This can be denoted H ! G.
• Task 3
Notation: For any group G and any subgroup H, G/H is the set of left cosets of H.
Formally, G/H = {gH | g is an element of G}.
We want to show that normality is a necessary and sufficient condition for G/H to be
a group.
In other words we will prove that G/H is a group and only if H is a normal subgroup
of G.
• Task 4
(Necessary: Step 1) Let H be a subgroup of G. Prove that if G/H is a group then
Hg = gH for every g in G.
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Hint: Let g be an element of G. What must you get if you multiply the cosets H and
gH? Now suppose that x is an element of Hg. . .
[A short technical proof - given next in Step 2 - will show that Hg ⊆ gH implies that
gH ⊆ Hg.]
• Task 5
(Necessary: Step 2) Let H be a subgroup of G.
We just proved that if G/H is a group then Hg ⊆ gH for all g ∈ G. Now we will show
that this in turn implies that gH ⊆ Hg for all g ∈ G. This will complete our proof
that if G/H is a group then H is a normal subgroup of G.
Proof. Suppose that G/H is a group. Let g be an element of G. We have shown that
Hg ⊆ gH.
Now suppose that x is an element of gH. Then x = gh for some h ∈ H. We wish to
show that x = gh = h! g for some h! ∈ H, so that we may conclude that gH ⊆ Hg.
Consider the equation gh = yg. Since G is a group, we know that there is a y that
satisfies this equation. Namely, y = ghg −1 . If we can show that y ∈ H then we will
have the desired result.
Note that hg −1 ∈ Hg −1 . We have previously shown that, for an arbitrary g, Hg ⊆ gH.
In particular, Hg −1 ⊆ g −1 H. Thus we can write hg −1 = g −1 h! for some h! ∈ H. Then
y = g(hg −1 ) = g(g −1 h! ) = h! and so y ∈ H as desired. Thus gH ⊆ Hg for any g ∈ G.!
Another way to see this, using the same coset containment argument Hg −1 ⊆ g −1 H,
involves everybody’s favorite algebra trick of multiplying by 1.
x =
=
=
=
=
=
gh
gh(g −1 g)
g(hg −1 )g
g(g −1 h! )g
(gg −1 )h! g
h! g
[multiply by identity]
[reassociate]
! −1
"
Hg ⊆ g −1 H
[reassociate]
Given our previous result, we now have that gH = Hg for all g ∈ G.
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