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A function f : AmapsB is said to be surjective (or onto) if for every y element of B there exists x element of A such that f(x) = y. Let G1 and G2 be groups. An isomorphism from G1 to G2 is an injection phi:G1mapsG2 such that phi(a · b) = phi(a) · phi(b)  (a,b element ofG1). A function f : AmapsB is said to be injective (or one to one) if whenever f(x) = f(y), we have x = y. If H is a subgroup of G then aH = {a · h | a in G, h in H} is a left cosetHa = {h · a | a in G, h in H} is a right coset The set of integers. Let H subset of G, H not equal empty set. Then H is a subgroup if and only if for a,belement ofH, we have a · b-1 element of H. f(a-1) = f(a)-1 (a element of G)

Homomorphisms

Recall that an isomorphism f from a group G into a group H is a function f :GmapsH which is injective and which satisfies

(1)
f(g · g') = f(g) · f(g')

for all g, g' element of G. There are many interesting examples of functions f :GmapsH which satisfy (1) but are not injective. Such functions are fundamental in the theory of groups. Let us make the following definition.

Definition 1: Let G and H be groups. A function f :GmapsH which satisfies (1) for all g, g' element of G is called a homomorphism of G into H.

Note that the product g · g' is taken with respect to multiplication in G, whereas the product f(g) · f(g') is taken with respect to multiplication in H.

We noted that isomorphic systems are essentially the same, being "abstractly equal". This is certainly not true of homomorphic systems in general. A homomorphic image can be compared to a silhouette of a system projected or mapped onto a screen. Certain features like the profile are preserved, but certain features such as coloration are lost.

Example 1: An isomorphism f :GmapsH is a homomorphism.

Example 2: Let f :GmapsH be defined by

f(g) = 1H.

for all g element of G. Then f is a homomorphism.

Example 3: Let G = Dn, H=[F]. Then every element of Dn is of the form FaRb (a = 0,1; b = 0,1,...,n-1). Let us now define f :GmapsH by

f(FaRb) = Fa

To prove that (1) holds, we must show that

(2)
f([FaRb] · [FcRd]) = f(FaRb) · f(FcRd).

However, we showed earlier that

[FaRb] · [FcRd] = Fa+cRb+d    if c = 0,
and
[FaRb] · [FcRd] = Fa+cRd-b    if c = 1.

Therefore, the left-hand side of (2) equals Fa+c in both cases, and

(3)
Fa+c = f(FaRb) · f(FcRd),

which proves (2). Thus, f is a homomorphism.

Example 4: Let f :ZmapsZ be defined by

f(a) = 2a    (a element of Z).

Since

f(a + b) = 2(a + b) = 2a + 2b = f(a) + f(b),

we see that f is a homomorphism.

Example 5: We may generalize Example 4 as follows: Let G be an abelian group, and let n be an integer. Then let us define fn :G mapsG by

fn(a) = an    (a element of G).

Since

fn(ab) = (ab)n = anbn = fn(a)fn(b),

we see that fn is a homomorphism.

Proposition 2: let f :GmapsH be a homomorphism.

(1) f(1G) = 1H.

(2) f(a-1) = f(a)-1    (a element of G).

(3) If K is a subgroup of G, then f(K) is a subgroup of H.

Proof: (1) Since 1G· 1G= 1G and since f is an isomorphism, we see that

f(1G) = f(1G· 1G) = f(1G)· f(1G).

Multiplying both sides by f(1G)-1, we get f(1G) = 1H.

(2) It suffices to show that

(4)
f(a-1)·f(a) = f(a)f(a-1) = 1H.

For then by the uniqueness of inverses in H, we have f(a-1) = f(a)-1. However,

a-1a = a·a-1 = 1G.

Therefore, by part (1) and the fact that f is a homomorphism, we get (4).

(3) By Proposition 2 of the section on subgroups, it suffices to show that if a,b element of K then f(a)· f(b)-1 element of f(K).

Suppose that f:GmapsH is a homomorphism. Set

ker(f) = {x element of G | f(x) = 1H}.

We will call ker(f) the kernel of f. If F is an isomorphism, then ker(f) = {1G}. The larger the kernel of f, the farther f is from an isomorphism.

If f:GmapsH is a homomorphism, then ker(f) is a normal subgroup of G.

Proposition 3: If f:GmapsH is a homomorphism, then ker(f) is a normal subgroup of G.

Proof: First let us show that ker(f) is a subgroup of G. Let a,b element of ker(f). It suffices to show that a · b-1 element of ker(f). Now

a,b element of ker(f) implies f(a) = f(b) = 1H
implies f(a) = f(b-1) = 1H (by part (2) of proposition 2 above.)
implies f(a · b-1) = f(a) f(b-1) = 1H
implies a · b-1 element of ker(f).

Let us now show that ker(f) is normal. Let x element of ker(f), a element of G. Then

f(axa-1) = f(a) · f(x) · f(a)-1
= f(a) · 1H· f(a)-1
=1H,

so that axa-1 element of ker(f). Therefore, ker(f) is normal.

If one knows the kernel of a homomorphism, then one has a good deal of information about the homomorphism, as the next two results show.

Let f :GmapsH be a homomorphism, h element of H. Let g element of G satisfy f(g) = h. Then f -1(h) = g · ker(f).

Proposition 4: Let f :GmapsH be a homomorphism, h element of H. Let g element of G satisfy f(g) = h. Then

f -1(h) = g · ker(f).

Proof: If g' element of g · ker(f), then g' = g · k for some k element of ker(f). Therefore,

f(g') = f(g · k) = f(g) f(k)
= f(g) [since k element of ker(f)]
= h.

Therefore, g' element of f -1 and g · ker(f) subset of f -1(h). In order to prove the reverse inclusion, note that if g' element of f -1(h), then f(g') = h implies f(g') · f(g)-1 = h · h-1 = 1H. But f(g')·f(g)-1 = f(g'·g-1) By Proposition 2 part(2). Therefore,

f(g'·g-1) = 1H implies g'·g-1 element of ker(f)
implies g' element of g · ker(f)

Thus, f -1(h) subset of g · ker(f), and therefore f -1(h) = g · ker(f).

Corollary 5: Let f :GmapsH be a homomorphism. Then f is an injection equivalent ker(f) = {1G}.

Proof: implies clear.

backwards implication Suppose that ker(f) = {1G. By Proposition 4, if h element of f(G), then f -1(h) consists of exactly one element. Therefore, f is an injection.

Proposition 3 tells us that with every homomorphism f of G into a group H there is an associated normal subgroup of G, ker(f). Conversely, suppose that we are given a normal subgroup N of G. Then there is an associated homomorphism, which we will now construct. We have already defined the quotient group G/N. Let us associate to N the homomorphism

iN:GmapsG/N,
iN(g) = gN.

Indeed, iN is a homomorphism since

iN(g1g2) = g1g2N
= g1N · g2N
= iN(g1)iN(g2)    (g1,g2 element of G)

by the definition of multiplication in G/N

Proposition 6: The homomorphism iN:GmapsG/N is surjective and ker(iN) = N.

Proof: It is clear that iN is surjective. Moreover,

g element of ker(iN iN(g) = 1G/N
gN = N
g element of N.

Therefore, ker(iN) = N.

The homomorphism iN:GmapsG/N is called the canonical homomorphism (associated with N). Note that by Proposition 6, every normal subgroup N of G appears as the kernel of a homomorphism of G, the canonical homomorphism.

Suppose that G, H, and J are groups, and suppose that a:GmapsH, b:HmapsJ, and c:GmapsJ are homomorphisms. It is easy to check the composite function ba:GmapsJ is also a homomorphism. We can picture this situation by drawing the following diagram:

homomorphism diagram

We see from the diagram that there are two homomorphisms defined from G into J, ba and c. If c = ba, then we day that the diagram is commutative. In other words, if the diagram is commutative, then the functions defined by the two paths from G to J are the same.

Let f :GmapsH be a homomorphism. Then we can construct the following diagram:

(5)
homomorphism diagram

Here i is the canonical homomorphism iker(f). A common sort of question which arises in many contexts is: Does there exist a homomorphism j:G/ker(f)mapsH [the dashed arrow in (5)] which makes the diagram commute? We will prove the existence and uniqueness of j in the following theorem.

Theorem 7 (1st [or fundamental] Isomorphism Theorem): Let f :GmapsH be a homomorphism. Then there exists a homomorphism j :G/ker(f)mapsH which makes diagram (5) commute. Such a j is unique. Moreover, j is an injection, so that j is an isomorphism and im(f) isomorphic to G/ker(f), where im(f) denotes the image of f. If, in addition, f is surjective, then j is surjective and H isomorphic to G/ker(f).

Remarks:

1. In many books, the first isomorphism theorem consists of the statement that if f :GmapsH is a surjective homomorphism, then H isomorphic to G/ker(f). Indeed, this is how the first isomorphism theorem is usually applied. However, for many purposes it is important to know not only that H and G/ker(f) are isomorphic, but that there is an isomorphism which makes the diagram (5) commute, which makes them isomorphic.

2. The real utility of Theorem 7 (as well as the other isomorphism theorems) is that they allow one to conclude that two groups are isomorphic to one another without actually constructing the isomorphism. Thus, these isomorphism theorems allow one to build a calculus for computing with groups.

3. The fundamental theorem asserts that G/ker(f) can be identified with the image of f under the isomorphism of j. Moreover, once this identification is carried out, the commutativity of the diagram implies that the homomorphism f :GmapsH is identified with the canonical homomorphism. Thus, roughly speaking, all homomorphisms f :GmapsH are "the same as" canonical homomorphisms. This is the real impact of the fundamental theorem.

Example 6: Let f :Gmaps{1G} be the homomorphism defined by f(g) = 1G for all g element of G. Then f is surjective and ker(f) = G, and from the theorem we conclude that G/G isomorphic to {1G}.

Example 7: Let f :Dnmaps[F] be the homomorphism defined by

f(Fa· Rb) = Fa.

Then f is surjective and ker(f) = [R]. Then from the theorem we conclude that Dn/[R] isomorphic to [F], a fact which could be easily verified directly.

Proof of Theorem: Let K = ker(f). Then a typical element of G/K is of the form gK (g element of G). Since i(g) = gK, if (5) commutes, then we must have

(6)
j(gK) = f(g).

Therefore, if j exists, it is unique and must be given by (6). Let us try to define j by (6). First, we must check that the definition of j(gK) depends only on the coset gK and not on the choice of g. But if gK = g'K, then gg'-1element of K, say gg'-1 = k, so that f(gg'-1) = 1H, since K = ker(f). But then, since f is a homomorphism, 1H = f(gg'-1) = f(g)f(g')-1, which implies that f(g) = f(g'). Therefore, the definition (6) makes sense. It is clear that j makes the diagram (5) commute. Moreover, j is a homomorphism since

j(gK·g'K) = j(gg'K) = f(gg') = f(g)f(g')
= j(gK)j(g'K).

Finally, j is an isomorphism, since if gK element of ker(f), then f(g) = 1H, which implies that g element of K and thus gK = K. Therefore, ker(j) = {1G/K}, and j is an isomorphism.