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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:GH 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:GG/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.

Theorem 11.10. First Isomorphism Theorem

If ψ:GH is a group homomorphism with K=kerψ, then K is normal in G. Let ϕ:GG/K be the canonical homomorphism. Then there exists a unique isomorphism η:G/Kψ(G) such that ψ=ηϕ.

Proof

We already know that K is normal in G. Define η:G/Kψ(G) by η(gK)=ψ(g). We first show that η is a well-defined map. If g1K=g2K, then for some kK, 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/Kψ(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(g11g2)=e, or g11g2 is in the kernel of ψ; hence, g11g2K=K; that is, g1K=g2K.

Mathematicians often use diagrams called commutative diagrams to describe such theorems. The following diagram “commutes” since ψ=ηϕ.

clipboard_ecc77b4458a2f370e66e65fed1b6d1165.png

Example 11.11

Let G be a cyclic group with generator g. Define a map ϕ:ZG by ngn.

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/mZG. 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.

Theorem 11.12

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, HN is a normal subgroup of H, and

H/HNHN/N.

Proof

We will first show that HN={hn:hH,nN} is a subgroup of G. Suppose that h1n1,h2n2HN. Since N is normal, (h2)1n1h2N. So

(h1n1)(h2n2)=h1h2((h2)1n1h2)n2

is in HN. The inverse of hnHN is in HN since

(hn)1=n1h1=h1(hn1h1).

Next, we prove that HN is normal in H. Let hH and nHN. Then h1nhH since each element is in H. Also, h1nhN since N is normal in G; therefore, h1nhHN.

Now define a map ϕ from H to HN/N by hhN. 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)=hhN=hNhN=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={hH:hN}=HN,

HN/N=ϕ(H)H/HN.

Theorem 11.13. Correspondence Theorem

Let N be a normal subgroup of a group G. Then HH/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)(b1N)=ab1NH/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={gG:gNS}, then for h1,h2H, we have that (h1N)(h2N)=h1h2NS and h11NS. Therefore, H must be a subgroup of G. Clearly, H contains N. Therefore, S=H/N. Consequently, the map HH/N is onto.

Suppose that H1 and H2 are subgroups of G containing N such that H1/N=H2/N. If h1H1, then h1NH1/N. Hence, h1N=h2NH2 for some h2 in H2. However, since N is contained in H2, we know that h1H2 or H1H2. Similarly, H2H1. Since H1=H2, the map HH/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/NG/H defined by gNgH 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

GG/NG/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.

Theorem \(11.14\. Third Isomorphism Theorem

Let G be a group and N and H be normal subgroups of G with NH. Then

G/HG/NH/N.

Example 11.15

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.


This page titled 11.2: The Isomorphism Theorms is shared under a GNU Free Documentation License 1.3 license and was authored, remixed, and/or curated by Thomas W. Judson (Abstract Algebra: Theory and Applications) via source content that was edited to the style and standards of the LibreTexts platform.

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