(vi) For all a, b ∈ G, ρ(a) = ρ(b) if and only if a ≡ b (mod ker(ρ)).

(vii) ρ is injective if and only if ker(ρ) = {0G }.

(viii) For any subgroup H of G , ρ’1 (H ) is a subgroup of G containing

ker(ρ).

Proof.

(i) We have

0G + ρ(0G ) = ρ(0G ) = ρ(0G + 0G ) = ρ(0G ) + ρ(0G ).

Now cancel ρ(0G ) from both sides (using part (i) of Theorem 8.3).

(ii) We have

0G = ρ(0G ) = ρ(a + (’a)) = ρ(a) + ρ(’a),

and hence ρ(’a) is the inverse of ρ(a).

(iii) For n = 0, this follows from part (i). For n > 0, this follows from

the de¬nitions by induction on n. For n < 0, this follows from the

positive case and part (v) of Theorem 8.3.

(iv) For any a, b ∈ H, we have a + b ∈ H and ’a ∈ H; hence, ρ(H)

contains ρ(a + b) = ρ(a) + ρ(b) and ρ(’a) = ’ρ(a).

196 Abelian groups

(v) If ρ(a) = 0G and ρ(b) = 0G , then ρ(a+b) = ρ(a)+ρ(b) = 0G +0G =

0G , and ρ(’a) = ’ρ(a) = ’0G = 0G .

(vi) ρ(a) = ρ(b) i¬ ρ(a) ’ ρ(b) = 0G i¬ ρ(a ’ b) = 0G i¬ a ’ b ∈ ker(ρ) i¬

a ≡ b (mod ker(ρ)).

(vii) If ρ is injective, then in particular, ρ’1 ({0G }) cannot contain any

other element besides 0G . If ρ is not injective, then there exist two

distinct elements a, b ∈ G with ρ(a) = ρ(b), and by part (vi), ker(ρ)

contains the element a ’ b, which is non-zero.

(viii) This is very similar to part (v). If ρ(a) ∈ H and ρ(b) ∈ H , then

ρ(a + b) = ρ(a) + ρ(b) ∈ H , and ρ(’a) = ’ρ(a) ∈ H . Moreover,

since H contains 0G , we must have ρ’1 (H ) ⊇ ρ’1 ({0G }) = ker(ρ).

2

Part (vii) of the above theorem is particular useful: to check that a group

homomorphism is injective, it su¬ces to determine if ker(ρ) = {0G }. Thus,

the injectivity and surjectivity of a given group homomorphism ρ : G ’ G

may be characterized in terms of its kernel and image:

• ρ is injective if and only if ker(ρ) = {0G };

• ρ is surjective if and only if img(ρ) = G .

The next three theorems establish some further convenient facts about

group homomorphisms.

Theorem 8.21. If ρ : G ’ G and ρ : G ’ G are group homomorphisms,

then so is their composition ρ —¦ ρ : G ’ G .

Proof. For a, b ∈ G, we have ρ (ρ(a + b)) = ρ (ρ(a) + ρ(b)) = ρ (ρ(a)) +

ρ (ρ(b)). 2

Theorem 8.22. Let ρi : G ’ Gi , for i = 1, . . . , n, be group homo-

morphisms. Then the map ρ : G ’ G1 — · · · — Gn that sends a ∈ G

to (ρ1 (a), . . . , ρn (a)) is a group homomorphism with kernel ker(ρ1 ) © · · · ©

ker(ρn ).

Proof. Exercise. 2

Theorem 8.23. Let ρi : Gi ’ G, for i = 1, . . . , n, be group homomor-

phisms. Then the map ρ : G1 — · · · — Gn ’ G that sends (a1 , . . . , an ) to

ρ1 (a1 ) + · · · + ρn (an ) is a group homomorphism.

Proof. Exercise. 2

Consider a group homomorphism ρ : G ’ G . If ρ is bijective, then ρ is

8.4 Group homomorphisms and isomorphisms 197

called a group isomorphism of G with G . If such a group isomorphism

ρ exists, we say that G is isomorphic to G , and write G ∼ G . Moreover,

=

if G = G , then ρ is called a group automorphism on G.

Theorem 8.24. If ρ is a group isomorphism of G with G , then the inverse

function ρ’1 is a group isomorphism of G with G.

Proof. For a , b ∈ G , we have

ρ(ρ’1 (a ) + ρ’1 (b )) = ρ(ρ’1 (a )) + ρ(ρ’1 (b )) = a + b ,

and hence ρ’1 (a ) + ρ’1 (b ) = ρ’1 (a + b ). 2

Because of this theorem, if G is isomorphic to G , we may simply say that

“G and G are isomorphic.”

We stress that a group isomorphism of G with G is essentially just a

“renaming” of the group elements ” all structural properties of the group

are preserved, even though the two groups might look quite di¬erent super-

¬cially.

Example 8.39. As was shown in Example 8.30, the quotient group G/H

discussed in that example is isomorphic to Z3 . As was shown in Exam-

ple 8.31, the quotient group Z— /(Z— )2 is isomorphic to Z2 — Z2 . As was

15 15

shown in Example 8.32, the quotient group Z— /(Z— )2 is isomorphic to Z2 . 2

5 5

Example 8.40. If gcd(n, m) = 1, then the m-multiplication map on Zn is

a group automorphism. 2

The following four theorems provide important constructions of group

homomorphisms.

Theorem 8.25. If H is a subgroup of an abelian group G, then the map

ρ : G ’ G/H given by ρ(a) = a + H is a surjective group homomorphism

whose kernel is H.

Proof. This really just follows from the de¬nition of the quotient group. To

verify that ρ is a group homomorphism, note that

ρ(a + b) = (a + b) + H = (a + H) + (b + H) = ρ(a) + ρ(b).

Surjectivity follows from the fact that every coset is of the form a + H for

some a ∈ G. The fact that ker(ρ) = H follows from the fact that a + H is

the coset of H in G containing a, and so this is equal to H if and only if

a ∈ H. 2

The homomorphism of the above theorem is called the natural map from

G to G/H.

198 Abelian groups

Theorem 8.26. Let ρ be a group homomorphism from G into G . Then

the map ρ : G/ ker(ρ) ’ img(ρ) that sends the coset a + ker(ρ) for a ∈ G

¯

to ρ(a) is unambiguously de¬ned and is a group isomorphism of G/ ker(ρ)

with img(ρ).

Proof. Let K := ker(ρ). To see that the de¬nition ρ is unambiguous, note

¯

that if a ≡ a (mod K), then by part (vi) of Theorem 8.20, ρ(a) = ρ(a ). To

see that ρ is a group homomorphism, note that

¯

ρ((a + K) + (b + K)) = ρ((a + b) + K) = ρ(a + b) = ρ(a) + ρ(b)

¯ ¯

= ρ(a + K) + ρ(b + K).

¯ ¯

It is clear that ρ maps onto img(ρ), since any element of img(ρ) is of the form

¯

ρ(a) for some a ∈ G, and the map ρ sends a + K to ρ(a). Finally, to see that

¯

ρ is injective, suppose that ρ(a + K) = 0G ; then we have ρ(a) = 0G , and

¯ ¯

hence a ∈ K; from this, it follows that a + K is equal to K, which is the zero

element of G/K. Injectivity then follows from part (vii) of Theorem 8.20,

applied to ρ. 2

¯

The following theorem is an easy generalization of the previous one.

Theorem 8.27. Let ρ be a group homomorphism from G into G . Then for

any subgroup H contained in ker(ρ), the map ρ : G/H ’ img(ρ) that sends

¯

the coset a + H for a ∈ G to ρ(a) is unambiguously de¬ned and is a group

homomorphism from G/H onto img(ρ) with kernel ker(ρ)/H.

Proof. Exercise ”just mimic the proof of the previous theorem. 2

Theorem 8.28. Let G be an abelian group with subgroups H1 , H2 . Then

the map ρ : H1 — H2 ’ H1 + H2 that sends (h1 , h2 ) to h1 + h2 is a surjective

group homomorphism. Moreover, if H1 © H2 = {0G }, then ρ is a group

isomorphism of H1 — H2 with H1 + H2 .

Proof. The fact that ρ is a group homomorphism is just a special case

of Theorem 8.23, applied to the inclusion maps ρ1 : H1 ’ H1 + H2 and

ρ2 : H2 ’ H1 + H2 . One can also simply verify this by direct calculation:

for h1 , h1 ∈ H1 and h2 , h2 ∈ H2 , we have

ρ(h1 + h1 , h2 + h2 ) = (h1 + h1 ) + (h2 + h2 )

= (h1 + h2 ) + (h1 + h2 )

= ρ(h1 , h2 ) + ρ(h1 , ρ2 ).

Moreover, from the de¬nition of H1 + H2 , we see that ρ is in fact surjective.