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ABSTRACT ALGEBRA 1, LECTURES NOTES 5: SUBGROUPS, CONJUGACY,
NORMALITY, QUOTIENT GROUPS, AND EXTENSIONS.
ANDREW SALCH
1. Subgroups, conjugacy, normality.
I think you already know what a subgroup is:
Definition 1.1. Let G be a group. A subgroup of G is a subset S of G satisfying the
following conditions:
• The unit element 1 of G is contained in S .
• If g1 , g2 ∈ S , then the product g1 g2 in G is also contained in S .
• If g ∈ S , then the inverse g−1 in G is also contained in S .
Definition 1.2. Let f : G → H be a group homomorphism. By the kernel of f , abbreviated
ker f , we mean the subset of G consisting of all elements g ∈ G such that f (g) = 1. By the
image of f , abbreviated im f , we mean the subset of H consisting of all elements of the
form f (g) for some g ∈ G.
Exercise 1.3. (Very easy.) Prove that, for every group G, a subset S of G is a subgroup of
G if and only if S is the image of a group homomorphism from some group to G.
The kernel of a group homomorphism is also a subgroup; I give a proof of this in
Theorem 2.5.
Definition 1.4. Let S be a subset of a group G, and let g ∈ G. By the conjugate of S by g,
written g−1 S g, we mean the subset {g−1 sg : s ∈ S } of G.
Given two subsets S , T of a group G, if there exists some element g ∈ G such that
g−1 S g = T , then we say that S is conjugate to T .
Definition 1.5. Let G be a group, and let H be a subgroup of G. We say that H is a normal
subgroup of G if, for all elements g ∈ G, we have g−1 Hg = H.
Since the subgroup {1} of G is always normal in G, and since G is always normal as a
subgroup of itself, we sometimes say that {1} and G are the trivial normal subgroups of G,
and when H is a normal subgroup of G which is not equal to {1} or G itself, we say that H
is a nontrivial normal subgroup of G.
If a group G has no nontrivial normal subgroups, we say that G is simple.
Another way to express the definition of normality is: a subgroup is normal iff the only
subgroup it is conjugate to is itself.
Observation 1.6. The following observation is very useful for making computations: suppose G is a group, S is a subset of G, and g, h ∈ G. Then (gh)−1 S (gh) = h−1 (g−1 S g)h,
i.e., the conjugate of S by gh is equal to the conjugate by h of the conjugate of S by g.
This observation is very elementary and it doesn’t look important until you actually start
checking whether subgroups are normal, and then this observation suddenly becomes very
Date: September 2015.
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ANDREW SALCH
useful, since it means that, if x−1 S x = S for all generators x of G, then g−1 S g = S for all
g ∈ G. So to check whether a subgroup of a group G is normal, you only have to check that
it remains fixed under conjugation by a set of generators of G.
Example 1.7. Let’s compute all the subgroups of the symmetric group Σ3 , and let’s see
which ones are conjugate to which other ones, and which ones are normal. Using the
presentation
hσ, τ | σ2 = id, τ3 = id, στ = τ2 σi
for Σ3 , we look at the subsets of Σ3 one by one, and see what subgroup of Σ3 each of them
generates. For example, if H is a subgroup of Σ3 containing τ, then the subgroup must contain 1, τ, and τ2 so that it is closed under multiplication in Σ3 , and if it additionally contains
σ or τσ, then (again, since it is closed under multiplication) it contains all the elements of
Σ3 . Using reasoning like this, we easily find the following list of all the subgroups of Σ3 :
(1) {1},
(2) {1, σ} Z/2Z,
(3) {1, τ, τ2 } Z/3Z,
(4) {1, τσ} Z/2Z,
(5) {1, τ2 σ} Z/2Z,
(6) {1, σ, τ, τσ, τ2 , τ2 σ} = Σ3 ,
a total of six subgroups of Σ3 . Which ones are conjugate to which other ones? Let’s see:
• First, the easiest conjugations: τ−1 {1}τ = {1},
• τ−1 Σ3 τ = Σ3 ,
• σ−1 {1}σ = {1},
• σ−1 Σ3 σ = Σ3 ,
• τ−1 {1, τ, τ2 }τ = {1, τ, τ2 },
• σ−1 {1, σ}σ = {1, σ}.
• For the remaining conjugations, we have to do at least a tiny bit of computation:
τ−1 στ = τσ, so τ−1 {1, σ}τ = {1, τσ}.
• τ−1 (τσ)τ = τ2 σ, so τ−1 {1, τσ}τ = {1, τ2 σ}.
• τ−1 (τ2 σ)τ = σ, so τ−1 {1, τ2 σ}τ = {1, σ}.
• σ−1 σσ = σ, so σ−1 {1, σ}σ = {1, σ}.
• σ−1 (τσ)σ = τ2 σ, so σ−1 {1, τσ}σ = {1, τ2 σ}.
• σ−1 (τ2 σ)σ = τσ, so σ−1 {1, τ2 σ}σ = {1, τσ}.
• σ−1 τσ = τ2 , so σ−1 {1, τ, τ2 }σ = {1, τ, τ2 }.
Here’s a way of picturing all this information: I will write 1, 2, 3, 4, 5, 6 for the six
subgroups listed above, with the same numbering I gave them above. Then σ and τ permute
the six subgroups, by conjugacy, by the arrows in the diagram:
(1.0.1)
1
τ
6
3
2_
4o
τ
σ
σ
τ
(
6 5.
I only drew in an arrow when there is a nontrivial action by σ or τ; so for example, I did
not draw any arrows between 1 and any other nodes in the diagram, since σ and τ both fix
the subgroup {1} under the conjugation action.
So Σ3 has exactly one nontrivial normal subgroup, namely {1, τ, τ2 }. (There is an easy
way to see that the subgroup {1, τ, τ2 } must be normal in Σ3 , by the way: if two finite
subgroups are conjugate, then you can easily see that they have to have the same number
ABSTRACT ALGEBRA 1, LECTURES NOTES 5: SUBGROUPS, CONJUGACY, NORMALITY, QUOTIENT GROUPS, AND EXTENSIONS.3
of elements. Since {1, τ, τ2 } is the only subgroup of Σ3 of order 3, it cannot be conjugate to
any other subgroup of Σ3 , so {1, τ, τ2 } must be normal in Σ3 .)
If you are using LaTeX and having difficulty typesetting diagrams like 1.0.1, let me
know and I will send you the LaTeX code for this diagram!
Note that the subgroups of Σ3 are of order 1, 2, 3, and 6, which are all divisors of 6, the
order of Σ3 . This is not a coincidence:
Exercise 1.8. (Lagrange.) Let G be a finite group and let H be a subgroup of G. Then the
order of H divides the order of G. (This is also problem 19 in section 1.7 of your textbook,
which also provides a hint you may want to look at.)
2. Quotient groups and extensions.
Definition-Proposition 2.1. Let G be a group and let H be a normal subgroup of G. Then
the quotient group G/H is the group defined as follows:
• The elements of G/H are the equivalence classes of symbols of the form gH, where
g ∈ G, under the equivalence relation given by letting g1 H be equivalent to g2 H if
g1 g−1
2 ∈ H. (The symbols gH are called cosets of H in G.)
• The multiplication is given by letting g1 H · g2 H = g1 g2 H.
Proof. I have to actually prove that G/H forms a group! This takes several steps:
• First, I need to show that the multiplication g1 H · g2 H = g1 g2 H on equivalence
classes is well-defined. Suppose that g1 H and g01 H are in the same equivalence
class, and suppose that g2 H and g02 H are in the same equivalence class. Then
g1 (g01 )−1 ∈ H and g2 (g02 )−1 ∈ H. I want to show that g1 g2 H is equivalent to
g01 g02 H, i.e., that (g1 g2 )−1 g01 g02 is in H. Since g1 (g01 )−1 ∈ H, we also know that
0 −1
−1
(g01 )−1 g1 = g−1
1 (g1 (g1 ) )g1 ∈ g1 Hg1 = H (here we are using the fact that H is
0 −1
normal!). Since g2 g2 ∈ H as well, their product (g01 )−1 g1 g2 (g02 )−1 is also in H.
Now here’s what happens if I conjugate that element by (g01 )−1 :
−1 (g01 )−1
(g01 )−1 g1 g2 (g02 )−1 (g01 )−1 = g1 g2 (g02 )−1 (g01 )−1
= g1 g2 (g01 g02 )−1
−1
∈ (g01 )−1 H(g01 )−1
= H,
again using the fact that H is normal. That’s what we wanted: g1 g2 (g01 g02 )−1 ∈ H
tells us that g1 g2 H is equivalent to g01 g02 H, so the multiplication I defined on G/H
is indeed well-defined, and doesn’t depend on what choices we make of cosets
representing each equivalence class. (Notice that I made use of the normality of
the subgroup H to do this; if H isn’t a normal subgroup of G, this multiplication
isn’t necessary well-defined!)
• Now I need to check that the multiplication I defined on G/H satisfies the axioms
to be a group. Clearly the equivalence class of 1H is the unit element of G/H. For
associativity, we have that
(g1 H · g2 H) · g3 H = g1 g2 H · g3 H
= (g1 g2 )g3 H
= g1 (g2 g3 )H
= g1 H · (g2 H · g3 H),
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ANDREW SALCH
so associativity of the multiplication on G/H follows easily from associativity of
the multiplication on G. Similarly, gH · g−1 H = 1H, so existence of inverses in G
easily implies existence of inverses in G/H.
Exercise 2.2. (Very easy.) Let G be a group and let H be a normal subgroup of G. Prove
that the function f : G → G/H sending g to gH is a surjective group homomorphism, and
prove that f is an isomorphism if and only if H = {1}.
Exercise 2.3. (Very easy.) Let G be a finite group and let H be a normal subgroup of G.
Prove that the order of G/H times the order of H is equal to the order of G.
Here is a little lemma which you probably already noticed is true:
Lemma 2.4. Let G, H be groups, and let f : G → H be a group homomorphism. (Recall
that this means that f (1) = 1 and f (g1 g2 ) = f (g1 ) f (g2 ) for all g1 , g2 ∈ G.) Then f (g−1 ) =
f (g)−1 for all g ∈ G.
Proof. For any g ∈ G,
f (g) f (g−1 ) = f (gg−1 )
= f (1)
= 1, and
f (g ) f (g) = f (g−1 g)
−1
= f (1)
= 1,
so f (g ) is the multiplicative inverse of f (g). So f (g−1 ) = f (g)−1 .
−1
The following theorem appears in section 3.3 of your textbook, and is often called the
“first fundamental isomorphism theorem”:
Theorem 2.5. Let G, H be groups and let f : G → H be a group homomorphism. Then
the kernel of f is a normal subgroup of G, and the quotient group G/ ker f is isomorphic
to the image of f .
Proof. Clearly, ker f contains the unit element of G, by the definition of a group homomorphism (it must send 1 to 1). If g1 , g2 ∈ ker f , then f (g1 g2 ) = f (g1 ) f (g2 ) = 1 · 1 = 1, so
g1 g2 ∈ ker f , so ker f is closed under multiplication. If g ∈ ker f , let g−1 be the inverse of
g in G; then, from Lemma 2.4, we know that f (g−1 ) = f (g)−1 = 1−1 = 1, so g−1 ∈ ker f .
So ker f is a subgroup of G.
To show that ker f is normal in G: let g ∈ G, and let x ∈ ker f . Then f (g−1 xg) =
f (g−1 ) f (x) f (g) = f (g)−1 · 1 · f (g) = 1, so g−1 xg ∈ ker f . So g−1 (ker f )g = ker f . So ker f
is normal.
Now let im f denote the image of f , and let h : G/ ker f → im f be defined by letting
h(g(ker f )) = f (g). (This notation is puzzling at a glance. What’s going on is that an
element in a quotient group G/H can be written as something of the form gH with g ∈ G,
so I can describe any element in G/ ker f as something of the form g(ker f ) with g ∈ G. So
g(ker f ) is something that it makes sense to apply h to.) (I leave the routine, easy exercise
of checking that h is well-defined to you, if you like to check these kinds of things.) I
need to show that h is injective and surjective. First, every element in im f is of the
form f (g) for some g ∈ G, hence f (g) = h(g(ker f )) for that same g, so h is surjective.
ABSTRACT ALGEBRA 1, LECTURES NOTES 5: SUBGROUPS, CONJUGACY, NORMALITY, QUOTIENT GROUPS, AND EXTENSIONS.5
Now suppose that g1 , g2 ∈ G and h(g1 (ker f )) = h(g2 (ker f )). Then f (g1 ) = f (g2 ), so
−1
1 = f (g1 ) f (g2 )−1 = f (g1 g−1
2 ), so g1 g2 ∈ ker f . So g1 (ker f ) is equivalent to g2 (ker f ), i.e.,
g1 (ker f ) and g2 (ker f ) represent the same element of G/ ker f . So h is injective. So h is an
isomorphism of groups.
Definition 2.6. Let G0 , G, G00 be groups, and let
(2.0.2)
f
g
G0 −→ G −→ G00
be group homomorphisms. We say that 2.0.2 is an extension of groups, also called a short
exact sequence of groups, if f is injective, g is surjective, and the image of f is equal to
the kernel of g. If an extension 2.0.2 exists, we say that G is an extension of G00 by G0 .
Exercise 2.7. (Easy.) Let 2.0.2 be an extension of groups. Prove that G0 is isomorphic to
a normal subgroup H of G, and that G00 is isomorphic to the quotient group G/H.
Exercise 2.8. Find all the extensions of groups 2.0.2, up to isomorphism, in which G = Σ3 .
(There is enough information in this set of lecture notes to allow you to do this easily.) Then
do the same for G = D8 .
One of the main purposes of extensions of groups is to give us a means of thinking of
complicated groups as being “built up” from smaller groups. For example, in the previous
exercise, you will find a short exact sequence which demonstrates that Σ3 is an extension of
an abelian group by an abelian group, so even though Σ3 is nonabelian, it is “built up”—via
an extension—from smaller groups which are both abelian. We will make much use of this
point of view in the next few weeks.