# First Isomorphism Theorem: Statement, Proof, Application

The first isomorphism theorem for groups proves that every homomorphic image of a group is actually a quotient group. This theorem is also known as the fundamental theorem of homomorphism. In this article, we will learn about the first isomorphism theorem for groups and the theorem is given below.

First isomorphism theorem of groups: Let G and G′ be two groups. If there is an onto homomorphism Φ from G to G′, then G/ker(Φ) ≅ G′.

## First Isomorphism Theorem for Groups

Statement: Let G and G′ be two groups. Let Φ: G → G′ be a homomorphism. Then we have the following isomorphism of groups

G/ker(Φ) ≅ Im(Φ)

## Proof of First Isomorphism Theorem

The below steps have to be followed to prove the first isomorphism theorem of groups. We will proceed as follows:

Step 1: We know that ker(Φ) is a normal subgroup of G. Thus the quotient group G/ker(Φ) is well-defined. So we define a map

θ: G/ker(Φ) → Im(Φ)

by θ(aH) = Φ(a) for all a ∈ G. Write H=ker(Φ). So θ: G/H → Im(Φ).

Let aH=bH. We will show that θ(aH)=θ(bH).

Since aH=bH we have a-1b ∈ H=ker(Φ). Thus we have

Φ(a-1b)=eG’

⇔ Φ(a)-1 Φ(b) = eG’

⇔ Φ(a) = Φ(b)

⇔ θ(aH)=θ(bH).

This shows that the map θ is well-defined and one-to-one.

Step 2: To show the map θ is a homomorphism. Write aH, bH ∈ G/H for a, b in G.

θ(aH ∗ bH) = θ(abH) where ∗ is the composition on the group G/H.

=Φ(ab)

=Φ(a) Φ(b) as Φ is a homomorphism

= θ(aH) θ(bH)

This shows that the map θ is a group homomorphism.

Step 4: By definition, the map θ is onto. This is because: take g’ ∈ Im(Φ); so we have g ∈ G such that Φ(g)=g’.

Now θ(gH) = Φ(g)=g’. So g’ has a preimage gH under the map θ. Hence, the map θ is onto.

Thus we have shown that the map θ is a one-to-one and onto homomorphism. In other words, θ is a group isomorphism, that is, G/H=G/ker(Φ) ≅ Im(Φ). This completes the proof of the first isomorphism theorem for groups.

Other Isomorphism Theorems:

Second Isomorphism Theorem

Third Isomorphism Theorem

## Properties of First Isomorphism Theorem

From the first isomorphism theorem of groups, we can conclude the following points:

• If G is a group, then every homomorphic image of G is isomorphic to a quotient group G/H for some normal subgroup H of G.
• In some sense, the natural quotient homomorphisms G→G/H describe all possible homomorphisms of G.
• The total number of homomorphisms of G = The number of different natural quotient homomorphisms G→G/H = The number of different normal subgroups H of G.
• The first isomorphism theorem for groups is also called the fundamental theorem of group homomorphisms.

## Application of First Isomorphism Theorem

We will give a few examples as an application of the first isomorphism theorem for groups.

Question 1: Let G be a group of order 12 and G′ be a group of order 5. Show that there does not exist a homomorphism from G onto G′.

Solution:

If possible, suppose that Φ: G → G′ is an onto homomorphism. Then by the first isomorphism of groups, we have that

G/ker(Φ) ≅ G′

Thus both groups G/ker(Φ) and G′ have the same order. So we get that

|G/ker(Φ)| = |G′|

⇒ |G| / |ker(Φ)| = |G′|

⇒ 12 / |ker(Φ)| = 5

⇒ |ker(Φ)| = 12/5 which is not a positive integer. Thus we arrive at a contradiction. So there does not exist a group homomorphism from G onto G′.

## FAQs on First Isomorphism Theorem

Q1: What is first isomorphism theorem?

Answer: The first isomorphism theorem for groups describes all the possible homomorphisms of a group G. It shows that every homomorphic image of G is actually a quotient group G/H for some choice of a normal subgroup H of G.

Q2: Can a cyclic group be isomorphic to a non-cyclic group?

Answer: As the cyclicity nature of group preserves under a group isomorphism, we conclude that a cyclic group can never be isomorphic to a non-cyclic group.

Share via: