11.2: The Isomorphism Theorms
( \newcommand{\kernel}{\mathrm{null}\,}\)
Although it is not evident at first, factor groups correspond exactly to homomorphic images, and we can use factor groups to study homomorphisms. We already know that with every group homomorphism ϕ:G→H we can associate a normal subgroup of G, kerϕ. The converse is also true; that is, every normal subgroup of a group G gives rise to homomorphism of groups.
Let H be a normal subgroup of G. Define the natural or canonical homomorphism
ϕ:G→G/H
by
ϕ(g)=gH.
This is indeed a homomorphism, since
ϕ(g1g2)=g1g2H=g1Hg2H=ϕ(g1)ϕ(g2).
The kernel of this homomorphism is H. The following theorems describe the relationships between group homomorphisms, normal subgroups, and factor groups.
If ψ:G→H is a group homomorphism with K=kerψ, then K is normal in G. Let ϕ:G→G/K be the canonical homomorphism. Then there exists a unique isomorphism η:G/K→ψ(G) such that ψ=ηϕ.
- Proof
-
We already know that K is normal in G. Define η:G/K→ψ(G) by η(gK)=ψ(g). We first show that η is a well-defined map. If g1K=g2K, then for some k∈K, g1k=g2; consequently,
η(g1K)=ψ(g1)=ψ(g1)ψ(k)=ψ(g1k)=ψ(g2)=η(g2K).
Thus, η does not depend on the choice of coset representatives and the map η:G/K→ψ(G) is uniquely defined since ψ=ηϕ. We must also show that η is a homomorphism. Indeed,
η(g1Kg2K)=η(g1g2K)=ψ(g1g2)=ψ(g1)ψ(g2)=η(g1K)η(g2K).
Clearly, η is onto ψ(G). To show that η is one-to-one, suppose that η(g1K)=η(g2K). Then ψ(g1)=ψ(g2). This implies that ψ(g−11g2)=e, or g−11g2 is in the kernel of ψ; hence, g−11g2K=K; that is, g1K=g2K.
Mathematicians often use diagrams called commutative diagrams to describe such theorems. The following diagram “commutes” since ψ=ηϕ.
Let G be a cyclic group with generator g. Define a map ϕ:Z→G by n↦gn.
Solution
This map is a surjective homomorphism since
ϕ(m+n)=gm+n=gmgn=ϕ(m)ϕ(n).
Clearly ϕ is onto. If |g|=m, then gm=e. Hence, kerϕ=mZ and Z/kerϕ=Z/mZ≅G. On the other hand, if the order of g is infinite, then kerϕ=0 and ϕ is an isomorphism of G and Z. Hence, two cyclic groups are isomorphic exactly when they have the same order. Up to isomorphism, the only cyclic groups are Z and Zn.
Let H be a subgroup of a group G (not necessarily normal in G) and N a normal subgroup of G. Then HN is a subgroup of G, H∩N is a normal subgroup of H, and
H/H∩N≅HN/N.
- Proof
-
We will first show that HN={hn:h∈H,n∈N} is a subgroup of G. Suppose that h1n1,h2n2∈HN. Since N is normal, (h2)−1n1h2∈N. So
(h1n1)(h2n2)=h1h2((h2)−1n1h2)n2
is in HN. The inverse of hn∈HN is in HN since
(hn)−1=n−1h−1=h−1(hn−1h−1).
Next, we prove that H∩N is normal in H. Let h∈H and n∈H∩N. Then h−1nh∈H since each element is in H. Also, h−1nh∈N since N is normal in G; therefore, h−1nh∈H∩N.
Now define a map ϕ from H to HN/N by h↦hN. The map ϕ is onto, since any coset hnN=hN is the image of h in H. We also know that ϕ is a homomorphism because
ϕ(hh′)=hh′N=hNh′N=ϕ(h)ϕ(h′).
By the First Isomorphism Theorem, the image of ϕ is isomorphic to H/kerϕ; that is,
HN/N=ϕ(H)≅H/kerϕ.
Since
kerϕ={h∈H:h∈N}=H∩N,
HN/N=ϕ(H)≅H/H∩N.
Let N be a normal subgroup of a group G. Then H↦H/N is a one-to-one correspondence between the set of subgroups H of G 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.
- Proof
-
Let H be a subgroup of G containing N. Since N is normal in H, H/N is a factor group. Let aN and bN be elements of H/N. Then (aN)(b−1N)=ab−1N∈H/N; hence, H/N is a subgroup of G/N.
Let S be a subgroup of G/N. This subgroup is a set of cosets of N. If H={g∈G:gN∈S}, then for h1,h2∈H, we have that (h1N)(h2N)=h1h2N∈S and h−11N∈S. Therefore, H must be a subgroup of G. Clearly, H contains N. Therefore, S=H/N. Consequently, the map H↦H/N is onto.
Suppose that H1 and H2 are subgroups of G containing N such that H1/N=H2/N. If h1∈H1, then h1N∈H1/N. Hence, h1N=h2N⊂H2 for some h2 in H2. However, since N is contained in H2, we know that h1∈H2 or H1⊂H2. Similarly, H2⊂H1. Since H1=H2, the map H↦H/N is one-to-one.
Suppose that H is normal in G and N is a subgroup of H. Then it is easy to verify that the map G/N→G/H defined by gN↦gH is a homomorphism. The kernel of this homomorphism is H/N, which proves that H/N is normal in G/N.
Conversely, suppose that H/N is normal in G/N. The homomorphism given by
G→G/N→G/NH/N
has kernel H. Hence, H must be normal in G.
Notice that in the course of the proof of Theorem 11.13, we have also proved the following theorem.
Let G be a group and N and H be normal subgroups of G with N⊂H. Then
G/H≅G/NH/N.
By the Third Isomorphism Theorem,
Z/mZ≅(Z/mnZ)/(mZ/mnZ).
Solution
Since |Z/mnZ|=mn and |Z/mZ|=m, we have |mZ/mnZ|=n.