# ABSTRACT ALGEBRA 1, LECTURES NOTES 5: SUBGROUPS, CONJUGACY, NORMALITY, QUOTIENT GROUPS, AND EXTENSIONS.

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1 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 g 1, g 2 S, then the product g 1 g 2 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

2 2 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 Σ 3, and let s see which ones are conjugate to which other ones, and which ones are normal. Using the presentation σ, σ 2 = id, 3 = id, σ = 2 σ for Σ 3, we look at the subsets of Σ 3 one by one, and see what subgroup of Σ 3 each of them generates. For example, if H is a subgroup of Σ 3 containing, then the subgroup must contain 1,, and 2 so that it is closed under multiplication in Σ 3, and if it additionally contains σ or σ, then (again, since it is closed under multiplication) it contains all the elements of Σ 3. Using reasoning like this, we easily find the following list of all the subgroups of Σ 3 : (1) {1}, (2) {1, σ} Z/2Z, (3) {1,, 2 } Z/3Z, (4) {1, σ} Z/2Z, (5) {1, 2 σ} Z/2Z, (6) {1, σ,, σ, 2, 2 σ} = Σ 3, a total of six subgroups of Σ 3. Which ones are conjugate to which other ones? Let s see: First, the easiest conjugations: 1 {1} = {1}, 1 Σ 3 = Σ 3, σ 1 {1}σ = {1}, σ 1 Σ 3 σ = Σ 3, 1 {1,, 2 } = {1,, 2 }, σ 1 {1, σ}σ = {1, σ}. For the remaining conjugations, we have to do at least a tiny bit of computation: 1 σ = σ, so 1 {1, σ} = {1, σ}. 1 (σ) = 2 σ, so 1 {1, σ} = {1, 2 σ}. 1 ( 2 σ) = σ, so 1 {1, 2 σ} = {1, σ}. σ 1 σσ = σ, so σ 1 {1, σ}σ = {1, σ}. σ 1 (σ)σ = 2 σ, so σ 1 {1, σ}σ = {1, 2 σ}. σ 1 ( 2 σ)σ = σ, so σ 1 {1, 2 σ}σ = {1, σ}. σ 1 σ = 2, so σ 1 {1,, 2 }σ = {1,, 2 }. 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 permute the six subgroups, by conjugacy, by the arrows in the diagram: (1.0.1) I only drew in an arrow when there is a nontrivial action by σ or ; so for example, I did not draw any arrows between 1 and any other nodes in the diagram, since σ and both fix the subgroup {1} under the conjugation action. So Σ 3 has exactly one nontrivial normal subgroup, namely {1,, 2 }. (There is an easy way to see that the subgroup {1,, 2 } must be normal in Σ 3, by the way: if two finite subgroups are conjugate, then you can easily see that they have to have the same number σ σ 3

3 ABSTRACT ALGEBRA 1, LECTURES NOTES 5: SUBGROUPS, CONJUGACY, NORMALITY, QUOTIENT GROUPS, AND EXTENSIONS.3 of elements. Since {1,, 2 } is the only subgroup of Σ 3 of order 3, it cannot be conjugate to any other subgroup of Σ 3, so {1,, 2 } must be normal in Σ 3.) 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 Σ 3 are of order 1, 2, 3, and 6, which are all divisors of 6, the order of Σ 3. 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 g 1 H be equivalent to g 2 H if g 1 g 1 2 H. (The symbols gh are called cosets of H in G.) The multiplication is given by letting g 1 H g 2 H = g 1 g 2 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 g 1 H g 2 H = g 1 g 2 H on equivalence classes is well-defined. Suppose that g 1 H and g 1H are in the same equivalence class, and suppose that g 2 H and g 2H are in the same equivalence class. Then g 1 (g 1 ) 1 H and g 2 (g 2 ) 1 H. I want to show that g 1 g 2 H is equivalent to g 1 g 2 H, i.e., that (g 1g 2 ) 1 g 1 g 2 is in H. Since g 1(g 1 ) 1 H, we also know that (g 1 ) 1 g 1 = g 1 1 (g 1(g 1 ) 1 )g 1 g 1 1 Hg 1 = H (here we are using the fact that H is normal!). Since g 2 g 1 2 H as well, their product (g 1 ) 1 g 1 g 2 (g 2 ) 1 is also in H. Now here s what happens if I conjugate that element by (g 1 ) 1 : ( (g 1 ) 1) 1 ( (g 1 ) 1 g 1 g 2 (g 2 ) 1) (g 1 ) 1 = g 1 g 2 (g 2 ) 1 (g 1 ) 1 = g 1 g 2 (g 1 g 2 ) 1 ( (g 1 ) 1) 1 H(g 1 ) 1 = H, again using the fact that H is normal. That s what we wanted: g 1 g 2 (g 1 g 2 ) 1 H tells us that g 1 g 2 H is equivalent to g 1 g 2H, 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 (g 1 H g 2 H) g 3 H = g 1 g 2 H g 3 H = (g 1 g 2 )g 3 H = g 1 (g 2 g 3 )H = g 1 H (g 2 H g 3 H),

4 4 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 (g 1 g 2 ) = f (g 1 ) f (g 2 ) for all g 1, g 2 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 1 ) f (g) = f (g 1 g) = f (1) = 1, so f (g 1 ) is the multiplicative inverse of f (g). So f (g 1 ) = f (g) 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 g 1, g 2 ker f, then f (g 1 g 2 ) = f (g 1 ) f (g 2 ) = 1 1 = 1, so g 1 g 2 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.

5 ABSTRACT ALGEBRA 1, LECTURES NOTES 5: SUBGROUPS, CONJUGACY, NORMALITY, QUOTIENT GROUPS, AND EXTENSIONS.5 Now suppose that g 1, g 2 G and h(g 1 (ker f )) = h(g 2 (ker f )). Then f (g 1 ) = f (g 2 ), so 1 = f (g 1 ) f (g 2 ) 1 = f (g 1 g 1 2 ), so g 1g 1 2 ker f. So g 1 (ker f ) is equivalent to g 2 (ker f ), i.e., g 1 (ker f ) and g 2 (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 G, G, G be groups, and let (2.0.2) G f G g G be group homomorphisms. We say that 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 exists, we say that G is an extension of G by G. Exercise 2.7. (Easy.) Let be an extension of groups. Prove that G is isomorphic to a normal subgroup H of G, and that G 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 = Σ 3. (There is enough information in this set of lecture notes to allow you to do this easily.) Then do the same for G = D 8. 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 Σ 3 is an extension of an abelian group by an abelian group, so even though Σ 3 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.

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