7 Homomorphisms and the First Isomorphism Theorem

Home , Homomorphism, Image (mathematics), Isomorphism, Kernel (algebra), Kernel (category theory), Kernel (set theory)

In each of our examples of factor groups, we not only computed the factor group but identified it as isomorphic to an already well-known group. Each of these examples is a special case of a very important theorem: the first isomorphism theorem. This theorem provides a universal way of defining and identifying factor groups. Moreover, it has versions applied to all manner of algebraic structures, perhaps the most famous being the rank–nullity theorem of linear algebra. In order to discuss this theorem, we need to consider two subgroups related to any group homomorphism.

7.1 Homomorphisms, Kernels and Images Deﬁnition 7.1. Let φ : G → L be a homomorphism of multiplicative groups. The kernel and image of φ are the sets

ker φ = {g ∈ G : φ(g) = eL} Im φ = {φ(g) : g ∈ G}

Note that ker φ ⊆ G while Im φ ⊆ L.

Similar notions The image of a function is simply its range Im φ = range φ, so this is nothing new. You saw the concept of kernel in linear algebra. For example if A ∈ M3×2(R) is a matrix, then we can deﬁne the linear map

2 3 LA : R → R : x 7→ Ax

In this case, the kernel of LA is precisely the nullspace of A. Similarly, the image of LA is the column- space of A. All we are doing in this section is generalizing an old discussion from linear algebra.

Theorem 7.2. Suppose that φ : G → L is a homomorphism. Then

1. φ(eG) = eL 2. ∀g ∈ G, (φ(g))−1 = φ(g−1)

3. ker φ / G

4. Im φ ≤ L

Proof. 1. Suppose that g ∈ G. Then

φ(g) = φ(geG) = φ(g)φ(eG) =⇒ eL = φ(eG)

by the left cancellation law.

2. Suppose that g ∈ G. Then

−1 −1 −1 −1 eL = φ(eG) = φ(gg ) = φ(g)φ(g ) =⇒ (φ(g)) = φ(g )

3. First suppose that k1, k2 ∈ ker φ. Then

φ(k1k2) = φ(k1)φ(k2) = eL =⇒ k1k2 ∈ ker φ

1 and

−1 −1 −1 φ(k1 ) = (φ(k1)) = eL =⇒ k1 ∈ ker φ It follows that ker φ is a subgroup of G. To see that ker φ is normal, recall, let g ∈ G and k ∈ ker φ, then

−1 −1 −1 φ(gkg ) = φ(g)φ(k)φ(g) = eL =⇒ gkg ∈ ker φ

This is one of the conditions equaivalent to ker φ / G.

4. Let φ(g1), φ(g2) ∈ Im φ. Then

φ(g1)φ(g2) = φ(g1g2) ∈ Im φ and

−1 −1 (φ(g1)) = φ(g1 ) ∈ Im φ Thus Im φ is a subgroup of L.

Examples

1. Recall that the trace function tr : Mn(R) → R is a homomorphism of additive groups. These are Abelian groups and so the kernel of tr is automatically normal without needing the above 1 Theorem. The additive group of trace-free matrices is a normal subgroup of (Mn(R), +):

ker tr = {A ∈ Mn(R) : tr A = 0} / Mn(R)

2. Let φ : Z36 → Z20 be deﬁned by φ(n) = 5n (mod 20). The kernel of φ is the subgroup ker φ = n : 5n ≡ 0 (mod 20)} = {0, 4, 8, 12, 16 / Z36

This is simply the cyclic group C5. ( 1 if σ even, 3. The map sgn : Sn → {1, −1} given by sgn(σ) = is a homomorphism of −1 if σ odd, groups. Here {1, −1} is a group under multiplication. Since the identity in the target group is 1, we have ker sgn = An, the alternating group of even permutations in Sn. Indeed An / Sn. × 4. det : GLn(R) → R is a homomorphism and so ker det = SLn(R) is a normal subgroup SLn(R) / GLn(R). Compare this with example 1. All these examples should suggest an idea to you. Not only is every kernel a normal subgroup, the converse is also true: any normal subgroup is the kernel of some homomorphism. This (loosely) is the 1st isomorphism Theorem to which we will come shortly. It may seem counter-intuiutive, but the homomorphism approach actually makes calculations with factor groups easier! As the next lemma shows, there is a very easy correspondence between the cosets of the kernel of a homomorphism, and the elements of the image.

1 This kernel is often written sln(R) (for the special-linear algebra), and is very much related to the special linear group A tr A SLn(R). Indeed the expression det e = e shows that matrix exponentiation is a map exp : sln(R) → SLn(R).

2 Lemma 7.3. Suppose that φ : G → L is a homomorphism. Then

g1 ker φ = g2 ker φ ⇐⇒ φ(g1) = φ(g2)

Proof. For all g1, g2 ∈ G, we have −1 g1 ker φ = g2 ker φ ⇐⇒ g2 g1 ∈ ker φ −1 ⇐⇒ φ(g2 g1) = eL −1 ⇐⇒ φ(g2) φ(g1) = eL

⇐⇒ φ(g1) = φ(g2) . The lemma tells us there is a bijective correspondence between the factor group G and the ker φ image Im φ. In the next theorem, we put this to use to help us determine what can possibly be a homomorphism. Theorem 7.4. Let φ : G → L be a homomorphism. 1. If G is a ﬁnite group then Im φ is a ﬁnite subgroup of L and its order divides that of G. More succinctly:

|G| < ∞ =⇒ |Im φ| |G|.

2. Similarly, |L| < ∞ =⇒ |Im φ| |L|. Note that we are only assuming one of the groups to be ﬁnite in each case.

Proof. 1. If G is a ﬁnite group, then Im φ is a ﬁnite subgroup of L. Lemma 7.3 tells us that

φ(g1) = φ(g2) ⇐⇒ g1 ker φ = g2 ker φ whence there are precisely as many elements of Im φ as there are distinct cosets of ker φ in G. But this is the deﬁnition of the index: |Im φ| = (G : ker φ) Given that G is ﬁnite, the discussion following Lagrange’s Theorem quickly says that

|G| = |Im φ| · |ker φ| =⇒ |Im φ| |G|

2. This is immediate from Lagrange’s Theorem: Im φ is a subgroup of a ﬁnite group L and so the order of Im φ divides the order of L.

Examples 1. Suppose that |G| = 17 and |L| = 13. How many distinct homomorphisms are there φ : G → L? If φ were such a homomorphism, the Theorem says that |Im φ| divides both |G| and |L|. But the only such positive integer is 1. Since the image of any homomorphism always contains the identity (φ(eG) = eL), it follows that there is only one homomorphism! φ must be the function deﬁned by φ(g) = eL, ∀g ∈ G. More generally, if gcd(|G| , |L|) = 1, then the only homomorphism φ : G → L is the trivial function φ : g 7→ eL.

3 2. How many homomorphisms are there φ : Z4 → S3? Again the Theorem tells us that |Im φ| divides 4 = |Z4| and 6 = |S3|: thus |Im φ| is either 1 or 2. If |Im φ| = 2 then Im φ is a subgroup of order 2 of S3. There are exactly 3 of these: {e, µi} for each i = 1, 2, 3. Since a homomorphism must map the identity to the identity we therefore have the homomorphisms

0  e  0  e  0  e  1 µ1 1 µ2 1 µ3 φ :   7→   , φ :   7→   , φ :   7→   . 1 2  e  2 2  e  3 2  e  3 µ1 3 µ2 3 µ3

Finally if Im φ has one element then φ is the trivial homomorphism φ(z) = e, ∀z ∈ Z4. There are therefore 4 distinct homomorphisms.

Theorem 7.5. There are exactly d = gcd(m, n) distinct homomorphisms φ : Zm → Zn, deﬁned by n φ(x) = k x (mod n) where k = 0, . . . , d − 1 d There are several ways of proving this: one is in the homework. We use the above discussion on the cardinality of the image.

Proof. Suppose φ : Zm → Zn is a homomorphism and d = gcd(m, n). Then |Im φ| d. However Im φ 2 is a subgroup of Zn which is necessarily cyclic. Recalling the structure of cyclic groups we see that n must be a subgroup of the unique cyclic subgroup of order d in Zn. This is generated by d . The only possible choices for our homomorphisms are therefore3 n φ (x) = k x (mod n) k d for each k = 0, 1, 2, . . . , d − 1. It remains only to check that these are well-deﬁned functions. For this, note that for any j ∈ Z we have, n n m n φ (x + jm) = k (x + jm) ≡ k x + knj ≡ k x (mod n) (since m ∈ Z) k d d d d d = φk(x)

Example Suppose that φ : Z12 → Z20 is a homomorphism. Since gcd(12, 20) = 4, the image of φ must be a subgroup of C4 = h5i ≤ Z20. There are four choices:

φ0(x) = 0, φ1(x) = 5x, φ2(x) = 10x, φ3(x) = 15x (mod 20)

Reversing the argument, we see that there are also four distinct homomorphisms ψ : Z20 → Z12, namely

ψ0(x) = 0, ψ1(x) = 3x, ψ2(x) = 6x, ψ3(x) = 9x (mod 12) 2 Recall that a cyclic group Zn has exactly one subgroup, which is itself cyclic, of each order d which divides n. Therefore if Im φ is a subgroup of Zn, then it is cyclic. Since its order divides d, it must also be a subgroup of the unique Cd ≤ Zn. 3 We may choose φk(1) for each k. If these are homomorphisms, then φk(2) = φk(1) + φk(1), etc.

4 7.2 The 1st Isomorphism Theorem We have seen that all kernels of group homomorphisms are normal subgroups. In fact all normal subgroups are the kernel of some homomorphism. We state this in two parts. Theorem 7.6 (1st Isomorphism Theorem). Let G be a group. . 1. Let H / G. Then γ : G → G deﬁned by γ(g) = gH is a homomorphism with ker γ = H. H . 2. If φ : G → L is a homomorphism with kernel H, then µ : G → Im φ deﬁned by µ(gH) = φ(g) is H an isomorphism. The Theorem can be summarised by the following formula, . G =∼ Im φ. ker φ

φ The notion of a commutative diagram is often useful for summarizing G / Im φ theorems. To say that a diagram commutes means that if there are < multiple paths from one side of the diagram to the other, they both γ µ ! give the same result. In this case one can travel from G to Im φ in two G. ways. For the 1st isomorphism theorem, this means φ = µ ◦ γ. ker φ . Deﬁnition 7.7. Given a normal subgroup H / G, the function γ : G → G : g 7→ gH is called the H canonical or fundamental homomorphism. Proof of Theorem. We check that the functions γ and µ have the properties we claim. 1. γ is certainly a well-deﬁned function: we need only check that it is a homomorphism with ker γ = H. However

γ(g1)γ(g2) = g1 H · g2 H = (g1g2)H = γ(g1g2) by the definition of multiplication in a factor group. Moreover, the identity in the factor group is H, whence ker γ = {g ∈ G : γ(g) = H} Thus g ∈ ker γ ⇐⇒ gH = H ⇐⇒ g ∈ H, whence the kernel of γ is H, as claimed. . 2. Clearly H / G (Theorem 7.2) and so the factor group G is well-defined. We need to check H . that µ : G → Im φ defined by µ(gH) = φ(g) is an isomorphism. H Well-definition and Bijectivity These are immediate from Lemma 7.3. . Homomorphism For all g H, g H ∈ G , we have 1 2 H

µ(g1 H · g2 H) = µ(g1g2 H) = φ(g1g2)

= φ(g1)φ(g2) (since φ is a homomorphism)

= µ(g1 H)µ(g2 H) µ is a well-deﬁned, bijective homomorphism and is therefore an isomorphism.

5 Examples Here are two examples where we can calculate all the pieces in the Theorem.

1. Let φ : Z10 → Z20 be the homomorphism with φ(1) = 4. Then, by the homomorphism prop- erty, φ(x) = 4x (mod 20). In particular

ker φ = {x ∈ Z10 : 4x ≡ 0 (mod 20)} = {0, 5} and,

Im φ = {4x (mod 20) : x ∈ Z10} = {0, 4, 8, 12, 16}

Observe that ker φ = h5i ≤ Z10 is isomorphic to C2 and Im φ = h4i ≤ Z20 is isomorphic to C5. The factor group is . n o Z10 = {0, 5}, {1, 6}, {2, 7}, {3, 8}, {4, 9} ker φ

The function {0, 5}  0  {1, 6}  4  Z .     µ : 10 → Im φ : {2, 7} 7→  8  i.e. µ(x + ker φ) = 4x (mod 20) ker φ     {3, 8} 12 {4, 9} 16

is the isomorphism from the Theorem.

2. Rather than starting with a homomorphism, suppose we simply wanted to identify the factor . group Z10 with a well-known group (here H = {0, 5} is our normal subgroup from before). H . We might search for a homomorphism ψ : Z → L whose kernel is H. Since Z10 clearly has 10 H 5 elements (5 cosets of H in Z10), the obvious thing to do is search for a homomorphism

ψ : Z10 → Z5 with ker ψ = H

It should be clear that ψ(x) = x (mod 5) is such a homomorphism! Since Im ψ = Z5, the ﬁrst . isomorphism theorem says that Z10 =∼ Z . The isomorphism µ in this case is simply H 5 µ(x + H) = x (mod 5)

The point of these examples is that the Theorem can be used in two ways, both of which help us ﬁnd alternative descriptions of factor groups. . • Start with a homomorphism φ : G → L and build an isomorphism µ : G =∼ Im φ. ker φ • Start with a normal subgroup H / G and construct a homomorphism φ with H = ker φ. Then . G =∼ Im φ. H

6 Direct product problems We quickly refresh the examples we saw in the last section in the language of the ﬁrst isomorphism theorem. . . 1. Given G = (Z4 × Z8) , deﬁne H h(0, 1)i

φ : Z4 × Z8 → Z4 : (x, y) 7→ x

This is clearly a homomorphism, with ker φ = {(0, 1)}. Indeed φ is also surjective. It follows that . (Z4 × Z8) ∼ = Im φ = Z4 h(0, 1)i

The explicit isomorphism is

µ : (x, y) + H 7→ x

This is precisely the inverse of the isomorphism ψ we stated when originally considering this example.

In the ﬁrst example, we pulled φ out of thin air, and it worked! In general, it is often a good idea to search for a homomorphism φ : G → G with ker φ = H: this typically gives you more room with wich to work. This is what we do in the next example. . . 2. For G = (Z4 × Z8) , we search for a homomorphism H h(0, 2)i

φ : Z4 × Z8 → Z4 × Z8

with ker φ = h(0, 2)i. It should seem reasonable to try4

φ(x, y) = (x, 4y)

Certainly (x, y) ∈ ker φ ⇐⇒ x = 0 ∈ Z4 and 4y = 0 ∈ Z8. This is if and only if 2 | y, whence ker φ = {(0, 2)}. The 1st isomorphism theorem says that . (Z4 × Z8) ∼ = Im φ = {(x, 4y) ∈ Z4 × Z8} h(0, 2)i

via the isomorphism

µ((x, y) + H) = (x, 4y)

The result is clearly isomorphic to Z4 × Z2 via (x, 4y) 7→ (x, y).

4We need φ(0, 2) = (0, 0). Moreover, if φ(1, 0) = (a, b) and φ(0, 1) = (c, d), then we must have φ(x, y) = (ax + cy, bx + dy), so any possible homomorphism must look like a linear map.

7 . . 3. We play the same game with G = (Z4 × Z8) , deﬁning H h(2, 4)i

φ : Z4 × Z8 → Z4 × Z8 : (x, y) 7→ (2x, y − 2x)

Certainly φ(x, y) = (0, 0) ⇐⇒ 2x = 0 ∈ Z4 and y = 2x ∈ Z8. Thus x = 2k is even, and y = 4k ∈ Z8 is the corresponding multiple of 4. We therefore have a homomorphism with ker φ = h(2, 4)i, whence . (Z4 × Z8) ∼ ∼ = Im φ = {(2x, y − 2x)} = {0, 2} × Z8 = Z2 × Z8 h(2, 4)i

Here are two further examples written in this language.

1. Let H = h2πi ≤ R and deﬁne φ : R → (C×, ·) by φ(x) = eix. Clearly φ is a homomorphism with

ker φ = {x ∈ R : eix = 1} = H

It follows that . R =∼ Im φ = S1 = {eix} H

Indeed µ(x + h2πi) = eix is the isomorphism guaranteed by the 1st isomorphism theorem.

2. Let φ : S1 → S1 : eiθ 7→ e2iθ. Clearly φ is a homomorphism with ker φ = {1, −1}. It follows that

1. S =∼ Im φ = S1 {1, −1}

In general, it can be very difﬁcult to construct a simple homomorphim with the correct kernel. The primary application of the theorem comes when starting with a given homomorphism.