1. Cosets
Cosets are arguably one of the strangest structures that students encounter in abstract algebra, along with factor groups, which are strongly related. Here’s a motivating question for this section: if H is a subgroup of a group G, then how are |H| and |G| related? A partial answer to this is contained in Lagrange’s Theorem.
Definition 1. Let G be a group and H a subgroup of G.A left coset of H with representative in G is the set gH = {gh : h ∈ H}. A right coset of H with representative in G is the set
Hg = {hg : h ∈ H}.
Warning. Cosets are NOT subgroups in general!
Example. Let K = {(1), (1 2)} in S3. The left cosets are (1)K = (1 2)K = {(1), (1 2)} (1 3)K = (1 2 3)K = {(1 3), (1 2 3)} (2 3)K = (1 3 2)K = {(2 3), (1 3 2)}
The right cosets are
K(1) = K(1 2) = {(1), (1 2)} K(1 3) = K(1 3 2) = {(1 3), (1 3 2)} K(2 3) = K(1 2 3) = {(2 3), (1 2 3)}
Note that, except for the coset of the elements in H, the left and right cosets are different.
Warning. In additive notation, we write g + H = {g + h : h ∈ H}. Note that additive groups are by definition abelian and so g + H = H + g.
Example. Let H = h3i = {0, 3} in Z6. The cosets are 0 + H = {0, 3} = 3 + H 1 + H = {1, 4} = 4 + H 2 + H = {2, 5} = 5 + H.
These notes are derived primarily from Abstract Algebra, Theory and Applications by Thomas Judson (16ed). Most of this material is drawn from Chapters 6, and 9-11. Last Updated: October 18, 2019 1 Example. Let L = {(1), (1 2 3), (1 3 2)} in S3. The left cosets are
(1)L = (1 2 3)L = (1 3 2)L = L (1 2)L = (1 3)L = (2 3)L = {(1 2), (1 3), (2 3)}
The right cosets are
L(1) = L(1 2 3) = L(1 3 2) L(1 2) = L(1 3) = L(2 3) = {(1 2), (1 3), (2 3)}
In this case, the left and right cosets are the same.
Lemma 1. Let H be a subgroup of G and suppose g1, g2 ∈ G. The following are equivalent.
(1) g1H ⊂ g2H
(2) g1H = g2H −1 −1 (3) Hg1 = Hg2 (4) g2 ∈ g1H −1 (5) g1 g2 ∈ H.
Proof. (1) ⇒ (2) Assume g1H ⊂ g2H. We claim the opposite inclusion holds. Let g2h ∈ g2H for 0 0 0 −1 some h ∈ H. Since g1 ∈ g1H ⊂ g2H, then g1 = g2h for some h . But then g2 = g1(h ) and so
0 −1 0 −1 g2h = g1(h ) h = g1((h ) h) ∈ g1H.
−1 −1 (2) ⇒ (3) Assume g1H = g2H. Let hg1 ∈ Hg1 with h ∈ H. Note that g1 ∈ g1H = g2H, so 0 0 −1 0 −1 −1 g1 = g2h for some h ∈ H. Then g1 = (h ) g2 and so
−1 0 −1 −1 0 −1 −1 −1 hg1 = h((h ) g2 ) = (h(h ) )g2 ∈ Hg2 .
−1 −1 Thus, Hg1 ⊂ Hg2 . A similar proof shows the reverse inclusion and the result follows. −1 −1 −1 −1 −1 −1 −1 (3) ⇒ (4) Assume Hg1 = Hg2 . Then g2 ∈ Hg2 = Hg1 so g2 = hg1 for some h ∈ H. −1 Thus, g2 = g1h ∈ g1H.
−1 (4) ⇒ (5) Assume g2 ∈ g1H. Then g2 = g1h for some h ∈ H. Thus, g1 g2 = h ∈ H. −1 −1 0 (5) ⇒ (1) Assume g1 g2 ∈ H. Let g1h ∈ g1H for some h ∈ H. By our assumption, g1 g2 = h for 0 0 −1 0 −1 some h ∈ H. Thus, g1 = g2(h ) and g1h = g2((h ) h) ∈ g2H. Therefore g1H ⊂ g2H.
Warning. Remember when we proved that the equivalence classes under some equivalence relation partition the set. This is when we need that result.
Theorem 2. Let H be a subgroup of a group G. The left cosets of H in G partition G. 2 Proof. Define an equivalence relation ∼ on G by g1 ∼ g2 if (and only if) g2 ∈ g1H. Clearly ∼ is reflexive because g1 = g1e ∈ g1H, so g1 ∼ g1. Suppose g1 ∼ g2. Then g2 ∈ g1H. By the previous lemma, this implies g1H = g2H and so g1 = g1e ∈ g1H = g2H. Thus, g2 ∼ g1 and so ∼ is reflexive.
Finally, suppose g1 ∼ g2 and g2 ∼ g3. Then g2 ∈ g1H and g3 ∈ g2H. Again using the above lemma, g1H = g2H and g2H = g3H, so g1H = g3H. Thus, g3 ∈ g1H so g1 ∼ g3.
The equivalence classes under this relation are the cosets and therefore a partition of G.
Note that the above result holds if we replace ‘left’ with ‘right’. One could prove it similarly or apply the lemma (exercise).
Definition 2. Let G be a group and H a subgroup. The index of H in G is the number of left cosets in G, denoted [G : H].
We will show momentarily that the number of left cosets is equal to the number of right cosets.
Example. In the previous examples, we have [Z6 : H] = 3, [S3 : K] = 3, and [S3 : L] = 2.
Theorem 3. Let H be a subgroup of G. The number of left cosets of H in G is the same as the number of right cosets.
Proof. Denote by LH the set of left cosets of H in G and by RH the set of right cosets of H in G.
We will establish a bijection between these sets, which shows that |LH | = |RH |. Define a map
φ : LH → RH gH 7→ Hg−1.
First we need to show that this map is well-defined. Suppose g1H = g2H. By the lemma, φ(g1H) = −1 −1 Hg1 = Hg2 = φ(g2H). Thus, φ is well-defined. −1 −1 Suppose φ(g1H) = φ(g2H), then Hg1 = Hg2 . Again, the lemma implies g1H = g2H, so φ is −1 −1 −1 injective. Let Hg ∈ RH . Then φ(g H) = H(g ) = Hg, so φ is surjective.
Thus, φ is bijective and so the result holds.
3 2. Lagrange’s Theorem
The next lemma proves a fact we observed in the examples.
Lemma 4. Let H be a subgroup of G. For all g ∈ G, |H| = |gH|.
Proof. Fix g ∈ G and define a map φ : H → gH by h 7→ gh. It is clear that φ is well-defined. We will show that it is bijective. It will then follow immediately that |H| = |gH|.
Suppose φ(h) = φ(h0) for h, h0 ∈ H. Then gh = gh0 and so by left cancellation, h = h0. Thus φ is injective. Next let gh ∈ gH for some h ∈ H, then φ(h) = gh so the map is surjective.
The proof of Lagrange’s Theorem is now simple because we’ve done the legwork already.
Theorem 5 (Lagrange’s Theorem). Let G be a finite group and H a subgroup of G. Then |G|/|H| = [G : H]. In particular, the order of H divides the order of G.
Proof. The group G is partitioned into [G : H] distinct left cosets. Each coset has exactly |H| elements by the previous lemma. Hence, |G| = |H|[G : H].
We’ll now examine a host of consequence of Lagrange’s Theorem.
Corollary 6. Suppose G is a finite group and g ∈ G. Then the order of g divides |G|, and g|G| = e.
Proof. Note that |g| = hgi. Thus, for the first statement we simply apply Lagrange’s Theorem with H = hgi. Now set d = |G|. Then |g| = k | d by the above Thus, d = k` for some integer `. Then d k` k ` ` g = g = (g ) = e = e.
Corollary 7. Let G be a group with |G| = p, p prime. Then G is cyclic and any g ∈ G, g 6= e, is a generator.
Proof. Let g ∈ G, g 6= e. Then |g| | |G| by the previous corollary, so |g| = 1 or p. But g 6= e so |hgi| > 1, so |hgi| = p. Thus, hgi = G.
Let’s take a moment to consider what we have just proved. The last corollary says that every group of prime power order is cyclic. Thus, if G is a group of order p, p prime, then G is essentially the same as Zp. We will formalize this in coming sections.
Corollary 8. Let H,K be subgroups of a finite group G such that K ⊂ H ⊂ G. Then [G : K] = [G : H][H : K].
Proof. We have by Lagrange’s Theorem, |G| |G| |H| [G : K] = = · = [G : H][H : K]. |K| |H| |K| 4 The converse of Lagrange’s Theorem is false in general. If n | |G|, this does not imply that there exists a subgroup H of G with |H| = n. For example, A4 has no subgroup of order 6. Now for a fun little diversion into number theory.
Definition 3. The Euler φ-function is defined as φ : N → N with φ(1) = 1 and
φ(n) = |{m ∈ N : 1 ≤ m < n and gcd(m, n) = 1}|.
This next theorem formalizes something we discussed much earlier.
Theorem 9. Let n ∈ N.
(1) |U(n)| = φ(n). (2) (Euler’s Theorem) Let a, n be integers such that n > 0 and gcd(a, n) = 1. Then aφ(n) ≡ 1 mod n. p−1 (3) (Fermat’s Little Theorem) Let p be any prime and suppose p - a. Then a ≡ 1 mod p. p Furthermore, for any b ∈ Z, b ≡ b mod p.
Proof. (1) By definition, |U(n)| is the set of integers m, 1 ≤ m < n that are invertible (under multiplication) mod n. We need only show that any such number is relatively prime to n. Suppose mk ≡ 1 mod n for some integer k. Then mk = n` + 1, or mk + n(−`) = 1 for some integer `. Thus, by the Euclidean Algorithm, gcd(m, n) = 1.
(2) Since |U(n)| = φ(n), then aφ(n) = 1 for all a ∈ U(n). This is equivalent to aφ(n) ≡ 1 mod n.
(3) Apply Euler’s Theorem with n = p.
5 3. Normal subgroups and factor groups
Warning. We are deviating from the structure of the book at this point. I make no apologies.
On one hand, normality encapsulates the idea of left and right cosets being the same, but more importantly they lead to the idea of factor groups. The motivation for this is simply that subgroups do not give a full sense of the structure of the groups. The factor groups give the complementary hidden structure.
Definition 4. A subgroup N of a group G is normal if gN = Ng for all g ∈ G.
Example. (1) If G is abelian, then every subgroup is normal.
(2) L = {(1), (1 2 3), (1 3 2)} is normal in S3.
(3) K = {(1), (1 2)} is not normal in S3.
Theorem 10. Let G be a group and N a subgroup. The following are equivalent.
(1) N is normal. (2) For all g ∈ G, gNg−1 ⊂ N. (3) For all g ∈ G, gNg−1 = N.
Proof. (1) ⇒ (2) Assume N is normal and fix g ∈ G. Let n ∈ N, then by the definition of normal, gN = Ng so gn = n0g for some n0 ∈ N. Thus, gng−1 = (n0g)g−1 = n0(gg−1) = n0 ∈ N, so gNg−1 ⊂ N.
(2) ⇒ (3) Assume gNg−1 ⊂ N for all g ∈ G. We claim the opposite inclusion holds (for all g ∈ G). Let n ∈ N and fix g ∈ G. By our assumption, g−1ng ⊂ N (since it holds for all g ∈ G). Thus, g−1ng = n0 for some n0 ∈ N, so n = gn0g−1 ∈ gNg−1. Therefore, N ⊂ gNg−1.
(3) ⇒ (1) Assume gNg−1 = N for all g ∈ G. We will show gN ⊂ Ng. The proof that Ng ⊂ gN is −1 0 0 0 similar. Let g ∈ G and n ∈ N. Then gng = n for some n ∈ N. Thus, gn = n g ∈ Ng.
Let N be a normal subgroup of a group G. We define a binary operation on the cosets by
(aN)(bN) = (ab)N.
Before we proceed, we should verify that this operation is actually well-defined. Said another way, we must verify that the product is independent of choice of coset representative.
Let aN = cN and bN = dN. We claim (aN)(bN) = (cN)(dN). Since c ∈ aN and d ∈ bN, then there exists n1, n2 ∈ N such that c = an1 and d = bn2. Thus
(cd)N = (an1)(bn2)N = (an1)b(n2N) = (an1)(bN) = (an1)(Nb) = a(n1N)b = aNb = (ab)N.
Note that bN = Nb by normality. 6 Warning. The operation for additive cosets is equivalent,
(a + N) + (b + N) = (a + b) + N.
We denote by G/N the set of cosets of N in G.
Theorem 11. Let N be a normal subgroup of G. The cosets of N in G form a group G/N (with the operation above) of order [G : N].
Proof. We have already shown that the operation is binary. We verify the group axioms. Through- out, let aN, bN, cN ∈ G/N. The operation is associative since
(aN)[(bN)(cN)] = (aN)(bcN) = (a(bc))N = ((ab)c)N = (abN)(cN) = [(aN)(bN)](cN).
For all aN ∈ G/N,(eN)(aN) = (ea)N = aN, so eN = N is the identity in G/N. Similarly, for all −1 −1 −1 aN ∈ G/N,(aN)(a N) = (aa )N = eN, so a N is the inverse of aN.
Definition 5. Let G be a group and N a normal subgroup. The group G/N is the factor group of G by N.
Example. Let G = S3 and N = {(1), (1 2 3), (1 3 2)}. (Note that N = A3). Then N is normal in G (we have already verified this) and the cosets are N and (1 2)N. The Cayley Table is given by N (1 2)N N N (1 2)N (1 2)N (1 2)N N
Observe that this table is equivalent to that of Z2.
Example. Consider the subgroup H = 3Z in Z. There are 3 cosets: 0 + H, 1 + H, 2 + H. Write out the Cayley Table for Z/H and observe how it is equivalent to that of Z3.
Example. Consider Dn generated by r, s with rn = id, s2 = id, srs = r−1.
−1 Let Rn be the subgroup of rotational symmetries. Then Rn is normal in Dn because srs = r ∈ Rn. Note that
|Dn/Rn| = [Dn : Rn] = |Dn|/|Rn| = 2n/n = 2.
Thus, the Cayley Table is equivalent to Z2.
7 4. Homomorphisms
Our next goal will be to explain how normal subgroups and factor groups arise naturally in the study of groups. In particular, a normal subgroups can be thought of as the kernel, or left over, part of a map between two groups.
Definition 6. A homomorphism is a function φ :(G, ·) → (H, ◦) between groups such that
φ(g1 · g2) = φ(g1) ◦ φ(g2) for all g1, g2 ∈ G.
The set im φ = {φ(g): g ∈ G} is called the image of φ.
One should think of a homomorphism as a structure preserving map, that is, it preserves the structure of the groups G and H. In fact, whenever you see the word morphism, this is generally what is meant.
× Example. (1) The function φ : GL2(R) → R given by φ(M) = det(M) is a homomorphism. Note that in this case φ is surjective but not injective.
(2) The function φ : M2(R) → R given by φ(M) = det(M) is not a homomorphism because det(M + N) 6= det(M) + det(N) in general.
(3) The function φ : Z2 7→ Z4 by φ(0) = 0 and φ(1) = 2. This map is injective but not surjective. n (4) Choose g ∈ G, G a group, and define φ : Z → G by φ(n) = g . Then φ is a homomorphism and im φ = hgi.
(5) Let C = ({1, −1}, ×). Define φ : Sn → C by σ(σ) = sgn σ where sgn σ = 1 if σ is even and −1 is σ is odd. (We say sgn σ is the sign of σ).
Proposition 12. Let φ : G → H be a homomorphism of groups.
(1) φ(eG) = eH . (2) For any g ∈ G, φ(g−1) = φ(g)−1. (3) If K is a subgroup of G, then φ(K) is a subgroup of H. (4) If L is a subgroup of H, then φ−1(L) = {g ∈ G : φ(g) ∈ L} is a subgroup of G. Furthermore, if L is normal in H then φ−1(L) is normal in G.
Proof. (1) Let g ∈ G, then
φ(g)eH = φ(g) = φ(geG) = φ(g)φ(eG).
By left cancellation, eH = φ(eG). (2) Let g ∈ G, then by (1), −1 −1 eH = φ(gg ) = φ(g)φ(g ). Hence, φ(g)−1 = φ(g−1). 8 (3) Let K be a subgroup of G. Since eG ∈ K, then K 6= ∅. Let a, b ∈ φ(K), so there exists −1 −1 k1, k2 ∈ K such that φ(k1) = a and φ(k2) = b. Then ab = φ(k1)φ(k2) ∈ K because K is a subgroup of G. Thus, φ(K) is a subgroup of H.
−1 −1 −1 (4) Let L be a subgroup of H. Since eG ∈ φ (L), then φ (L) 6= ∅. Choose c, d ∈ φ (L), so φ(c), φ(d) ∈ L. Then by (2) φ(cd−1) = φ(c)φ(d−1) = φ(c)φ(d)−1 ∈ L because L is a subgroup of H. Thus, cd−1 ∈ φ−1(L) and so φ−1(L) is a subgroup of G.
Finally, suppose L is normal in H. Let g ∈ H. We claim gφ−1(L)g−1 ∈ φ−1(L). Let a ∈ φ−1(L), then φ(gag−1) = φ(g)φ(a)φ(g)−1 ∈ L because L is normal. Thus, gag−1 ∈ φ−1(L) for all a ∈ −1 −1 φ (L), so φ (L) is normal in G.
Definition 7. The kernel of a homomorphism φ : G → H is ker φ = {g ∈ G : φ(g) = eH }.
Theorem 13. Let φ : G → H be a group homomorphism. Then ker φ is a normal subgroup of G.
−1 Proof. The trivial group {eH } is a normal subgroup of H and ker φ = φ ({eH }). The result now follows from Proposition 12 (4).
× Example. (1) φ : GL2(R) → R given by φ(M) = det(M). Then ker φ = SL2(R). (2) φ : Z2 7→ Z4 by φ(0) = 0 and φ(1) = 2. We can check directly that ker φ = {0, 2}. n (3) Choose g ∈ G, G a group, and define φ : Z → G by φ(n) = g . If |g| = ∞ then ker φ = {0}. If |g| = k, then ker φ = kZ. (4) Let φ : Sn → C by σ(σ) = sgn σ. Then ker φ = An.
Example. Let G be a group and N a normal subgroup. The quotient map φ : G → G/N is given by φ(g) = gN. This is a homomorphism since for all g, h ∈ G,
φ(g)φ(h) = (gN)(hN) = (gh)N = φ(gh).
The quotient map is always surjective (the preimage of gN is g). The kernel of the quotient map is ker φ = N and so the quotient map is injective if and only if N is the trivial group.
The previous example illustrates the idiom: Kernels are normals and normals are kernels.
9 5. Isomorphisms
Definition 8. Two groups (G, ·) and (H, ◦) are said to be isomorphic if there exists a bijective homomorphism φ : G → H. The map φ in this case is called an isomorphism.
Example. (1) Let H be a cyclic group of order 3 generated by h (so H = {e, h, h2}). The map
φ : Z3 → H is an isomorphism. Note that we have already verified that this map is an isomorphism, we need only verify that it is injective and surjective, but this is an easy exercise in this case.
(2) The groups D3 and S3 are isomorphic. One possible map is φ : D3 → S3 given by setting φ(r) = (123) and φ(s) = (12).
Theorem 14. Let φ : G → H be an isomorphism of groups.
(1) φ−1 : H → G is an isomorphism. (2) |G| = |H|. (3) If G abelian, then H is abelian. (4) If G is cyclic, then H is cyclic. (5) If G has a subgroup of order n, then H has a subgroup of order n.
Proof. (1) By standard results on functions, φ−1 exists and is bijective (see the book’s introductory chapter). We need only verify that it is a homomorphism, but this is easy. Since φ(ab) = φ(a)φ(b) for all a, b ∈ G, then φ−1(ab) = φ−1(a)φ−1(b) for all a, b ∈ G.
(2) This follows because φ is bijective.
(3) Assume G is abelian. Let h1, h2 ∈ H, then there exist (unique) elements g1, g2 ∈ G such that
φ(g1) = h1 and φ(g2) = h2. Then
h1h2 = φ(g1)φ(g2) = φ(g1g2) = φ(g2g1) = φ(g2)φ(g1) = h2h1. Thus, H is abelian.
(4) Exercise.
(5) Let K be a subgroup of G of order n. Then φ(K) is a subgroup of H because φ is a homomor- phism. Moreover, |φ(K)| = n because φ is bijective. Proposition 15. A surjective group homomorphism φ : G → H is an isomorphism if and only if ker φ = {eH }.
Proof. Assume φ is an isomorphism. Because it is a homomorphism, eH ∈ ker φ. Because it is 1-1, then ker φ = {eH }.
Assume ker φ = {eH }. We claim φ is 1-1. Suppose φ(g1) = φ(g2) for some g1, g2 ∈ G. Then −1 −1 −1 e = φ(g1)φ(g2) = φ(g1g2 ). Thus, by our assumption, g1g2 = e so g1 = g2. It follows that G is 1-1 and thus an isomorphism. 10 The next theorem may be regarded as a classification of all cyclic groups (up to isomorphism).
Theorem 16. Let G be a cyclic group with generator a ∈ G. ∼ (1) If |a| = ∞, then G = Z. ∼ (2) If |a| = n < ∞, then G = Zn.
n Proof. Define a map φ : Z → G by φ(n) = a . Note that we have already verified that φ is a homomorphism. We verify in (1) that φ is bijective and leave (2) as an exercise.
It is clear that φ is surjective. We will show that it is injective and thus an isomorphism. Let k ` k−` k, ` ∈ Z such that φ(k) = φ(`). Then a = a and so a = e. Since |a| = ∞, this implies k − ` = 0, so k = `. Thus, φ is injective and the result follows. ∼ Corollary 17. If |G| = p, p prime, then G = Zp.
Proof. Let a ∈ G, a 6= e. Then by Lagrange’s Theorem, |a| | p. Since |a| 6= 1 then |a| = p. Thus, ∼ G = hai. By the previous theorem, G = Zp.
The next result tells us that every group is (isomorphic to) a subgroup of Sn.
Theorem 18 (Cayley’s Theorem). Every group is isomorphic to a group of permutations.
Proof. Let G be a group and for each g ∈ G define a map λg : G → G by λg(a) = ga for all −1 −1 a ∈ G. Note that if h ∈ G, then λg(g h) = g(g h) = h, so λg is surjective. If a, b ∈ G such that 1 λg(a) = λg(b), then ga = gb so a = b by left cancellation. Thus, λg is injective and hence bijective
It follows that λg is a permutation of (the set) G.
Now set G = {λg : g ∈ G}. We claim that G is a group under function composition. Of course, function composition is associative so we need only check closure under the operation, existence of an identity, and closure under inverses. First claim that for λgλh = λgh for all g, h ∈ G. These are functions, so it suffices to check this by evaluating both sides at some a ∈ G:
λgh(a) = (gh)a = g(ha) = g(λha) = λg(λha) = (λgλh)(a).
Now it is clear that λe is the identity of G and that λg−1 is the inverse of λg for all g ∈ G. Thus, G is a group.
Finally, we define a map Φ : G → G by g 7→ λg. By the previous paragraph, Φ is a homomorphism. We claim that Φ is an isomorphism. It is clear that Φ is surjective. For injectivity, let g, h ∈ G such that Φ(g) = Φ(h), so λg = λh. Now if a ∈ G, then ga = λg(a) = λh(a) = ha, so g = h by right cancellation. Thus, Φ is bijective and further an isomorphism.
1 Note that λg is not a homomorphism in general. 11 6. Direct products
Recall that if (G, ·) and (H, ◦) are groups, then G × H = {(g, h): g ∈ G, h ∈ H} is a group under the operation ? below,
(g1, h1) ? (g2, h2) = (g1 · g2, h1 ◦ h2) for all g1, g2 ∈ G, h1, h2 ∈ H. ∼ Recall that if |G| = p, p prime, then G = Zp. Here is a related result. The proof is left as an exercise.
∼ Proposition 19. Let m, n be positive integers, then Zm × Zn = Zmn if and only if gcd(m, n) = 1.
Definition 9. If G is a group and A, B subgroups of G such that G =∼ A × B, then G is said to be the external direct product of A and B.
∼ ∼ Example. The cyclic group Z6 has subgroups A = h2i = Z3 and B = h3i = Z2. By the above ∼ proposition, Z6 = Z2 × Z3. Thus, Z6 is the external direct product of A and B.
For a group G with subgroups H,N, we define the set
HN = {hn : h ∈ H, n ∈ N}.
Note that in general this is not a subgroup of G. The next proposition gives a criteria where it is.
Proposition 20. Let G be a group with subgroups H,N. If N is normal, then HN is a subgroup of G.
−1 Proof. Clearly e ∈ HN because e ∈ H ∩ N. Let h1n1, h2n2 ∈ HN. Then (h1n1)(h2n2) = −1 −1 (h1n1)(n2 h2). Because N is normal, Nh2 = h2N. Thus, n2 h = hn3 and n1h = hn4 for some n3, n4 ∈ N. We therefore have
−1 −1 (h1n1)(h2n2) = (h1n1)(n2 h2) = (h1n1)(h2n3) = h1(h2n4)n3 = (h1h2)(n4n3) ∈ HN.
Thus, HN is a subgroup of G.
Warning. In additive notation, HN is H + N = {h + n : h ∈ H, n ∈ N}. Recall that subgroups of abelian groups are always normal and so H + N is always a subgroup of G.
Exercise. Let H,N be subgroups of a group G with N normal. Prove that N is normal subgroup of HN and H ∩ N is a normal subgroup of H.
Definition 10. A group G is the interal direct product of subgroups H and K provided
(1) G = HK (as sets), (2) H ∩ K = {e}, and (3) hk = kh for all h ∈ H, k ∈ K. 12 Example. Let G = U(8), H = {1, 3}, K = {1, 5}. Then G = HK (because 3 · 5 ≡ 7 mod 8). Clearly H ∩ K = {1}, and because G is abelian we have hk = kh for all h ∈ H, k ∈ K. Thus, G is the internal direct product of H and K.
Example. S3 is not an internal direct product of its subgroups. Since |S3| = 6 then without loss ∼ ∼ of generality we need subgroups H,K with |H| = 3 and |K| = 2. But then H = Z3 and K = Z2. The condition hk = kh for all h ∈ H, k ∈ K now implies that S3 is abelian, a contradiction.
Lemma 21. If G is the internal direct product of subgroups H and K, then every element in G can be written uniquely as hk for some h ∈ H, k ∈ K.
Proof. Let g ∈ G. By the first condition of internal direct products, G = HK, and so g = hk for some h ∈ H, k ∈ K. We claim this decomposition is unique. Suppose g = h0k0 for some h0 ∈ H, k0 ∈ K. By the second condition, hk = h0k0 and so (h0)−1h = k0k−1 ∈ H ∩ K = {e}. Thus, 0 −1 0 −1 0 0 (h ) h = e and k k = e, so h = h and k = k.
The next theorem says that if G is the internal direct product of subgroups H and K, then it is also the external direct product of those subgroups.
Theorem 22. If G is the internal direct product of subgroups H and K, then G =∼ H × K.
Proof. Define a map φ : H × K → G by (h, k) 7→ hk. By the lemma, φ is bijective. We need only show that it is a homomorphism. Let (h1, k1), (h2, k2) ∈ H × K. Then by the third condition of internal direct products,
φ((h1, k1)(h2, k2)) = φ(h1h2, k1k2) = (h1h2)(k1k2) = (h1k1)(h2k2) = φ(h1, k1)φ(h2, k2).
Thus, φ is a homomorphism and therefore an isomorphism. ∼ Example. By a previous example, U(8) = Z2 × Z2.
Here is a result that is helpful in checking the internal direct product properties above.
Proposition 23. Let H and K be subgroups of a group G. Then |HK| = |H||K|/|H ∩ K|.
Proof. Let C = H ∩ K and note that C is a subgroup of K. By Lagrange’s Theorem n = [K : C] =
|K|/|H ∩ K|. Thus, K is the disjoint union of n right cosets: Ck1 ∪ Ck2 ∪ · · · Ckn for ki ∈ K. But
HC = C so then HK is the disjoint union Hk1 ∪ Hk2 ∪ · · · Hkn. Thus,
|HK| = |H| · n = |H||K|/|H ∩ K|.
Example. Let G = Zmn with gcd(m, n) = 1. Set H = hmi and K = hni so |H| = n and |K| = m. Clearly hk = kh for all h ∈ H, k ∈ K because G is abelian. Let g ∈ H ∩ K, then |g| divides n and m by Lagrange’s Theorem. But then |g| = 1 because gcd(m, n) = 1. Thus, g = 0 and so H ∩ K = {0}. Hence, by the previous proposition |HK| = |H||K| = nm = |G|, so G = HK.
13 7. The isomorphism theorems
The following theorems explain how factor group structures fit into the overall picture of group structure. They are also incredibly powerful tools for proving that two groups are isomorphic.
Our next result actually generalizes an earlier result that a homomorphism is an isomorphism if and only if its kernel is trivial. The first isomorphism theorem says that the factor group of a group by the kernel of an homomorphism is isomorphic to the image of the homomorphism.
Theorem 24 (First Isomorphism Theorem). Let φ : G → H be a homomorphism and K = ker φ. There exists an isomorphism
ψ : G/K → φ(G) gK 7→ φ(g).
That is, G/K =∼ φ(G).
Proof. First recall that by an earlier exercise, K is a normal subgroup and thus G/K is well-defined. However, ψ is a map defined on a set of cosets and therefore we must check that it is well-defined.
Suppose g1K = g2K for some g1K, g2K ∈ G/K. We must show that ψ(g1K) = ψ(g2K). Because −1 −1 K is normal, our hypothesis implies that g2 g1 ∈ K. Therefore, ψ(g2 g1) = e, so ψ(g2) = φ(g1). But then ψ(g1K) = φ(g2K). Thus, ψ is well-defined.
We now check that ψ is a homomorphism. Let g1K, g2K ∈ G/K. Then
ψ((g1K)(g2K)) = ψ((g1g2)K) = φ(g1g2) = φ(g1)φ(g2) = ψ(g1K)ψ(g2K).
It is left to show that ψ is bijective. Clearly ψ is surjective since for all φ(g) ∈ φ(G), ψ(gK) = φ(g).
To show injectivity we reverse the argument above for well-definedness. Suppose ψ(g1K) = ψ(g2K) −1 for some g1K, g2K ∈ G/K. Then φ(g1) = φ(g2) so g2 g1 ∈ K. Because K is normal, g1K = g2K, so ψ is injective.
Example. Consider the map φ : Z 7→ Z3 given by φ(k) = k mod 3. This map is a surjective ∼ homomorphism (check!) and ker φ = 3Z. By the first isomorphism theorem, Z/3Z = Z3. ∼ Exercise. Generalize the previous example (and carefully check all steps) to show that Z/nZ = Zn for all positive integers n.
Theorem 25 (Second Isomorphism Theorem). Let H,N be subgroups of a group G with N normal. Then H/H ∩ N =∼ HN/N.
Proof. By an earlier exercise, N is a normal subgroup of HN and H ∩ N is a normal subgroup of H. Thus, the given factor groups are well-defined. 14 Define a map
φ : H → HN/N h 7→ hN.
We claim this map is a surjective homomorphism. Let h1, h2 ∈ H. Then by the normality of N,
φ(h1h2) = (h1h2)N = h1(h2N) = h1(Nh2) = (h1N)(Nh2) = (h1N)(h2N) = φ(h1)φ(h2). Thus, φ is a homomorphism. Let hnN ∈ HN/N, then clearly hnN = hN = φ(h), so φ is surjective.
By the first isomorphism theorem, H/ ker φ =∼ HN/N, so we need only show that ker φ = H ∩ N. Let x ∈ H ∩ N, then φ(x) = xN = N because x ∈ N, so H ∩ N ⊂ ker φ. On the other hand, if y ∈ ker φ, then yN = φ(y) = N, so y ∈ N. Thus, ker φ ⊂ H ∩ N so the sets are equal.
Warning. In additive notation, the theorem is rephrased as H + N instead of HN.
Example. Let H = 2Z and N = 3Z be subgroups of Z. Clearly N is normal because Z is abelian. ∼ Then H ∩ N = 6Z and H + N = Z. Thus, by the second isomorphism theorem, 2Z/6Z = Z3.
The next two results are also important, but in the interest of time and sanity I will leave the proofs as reading exercises. However, I suspect that the third isomorphism theorem is one that you can prove once you fully understand the proofs above, especially that of the second isomorphism theorem.
Theorem 26 (Correspondence Theorem). Let N be a normal subgroup of a group G. Then there is a bijection from the set of all subgroups containing N and the set of subgroups of G/N. Furthermore, the normal subgroups of G containing N correspond to normal subgroups of G/N.
∼ Example. Consider the group G = Z12 and let N = h4i. Note that G/N = Z4. Then we can identify the subgroups of G containing N with the subgroups of G/N.
subgroups of G containing N 7→ subgroups of G/N N 7→ {N} {0, 2, 4, 6, 8, 10, 12} 7→ {N, 2 + N} G 7→ {N, 1 + N, 2 + N, 3 + N}.
Theorem 27 (Third Isomorphism Theorem). Let H,N be normal subgroups of a group G with N ⊂ H. Then G/H =∼ (G/N)/(H/N). ∼ Example. By the third isomorphism theorem, Z/3Z = (Z/6Z)/(3Z/6Z)
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