An isomorphism of groups is a special kind of group homomorphisms. It preserves every structure of groups. In this article, we will learn about isomorphism between groups, related theorems, and applications.

Table of Contents

## Definition of Isomorphism

A map Φ: (G, 0) → (G′, *) between two groups is called an isomorphism if the following conditions are satisfied:

- Φ is a group homomorphism, that is, Φ(ab)=Φ(a)Φ(b) ∀ a, b ∈ G.
- Φ is one-to-one.
- Φ is onto.

A bijective group homomorphism between groups is called an isomorphism.

For example, the identity map i: Z → Z defined by i(n)=n ∀ n ∈ Z is an example of an isomorphism. Below are a few more examples of isomorphism of groups.

**•** The map Φ: (Z_{5}, +) → (Z_{5}, +) defined by Φ($\bar{x}$)=3$\bar{x}$ ∀ $\bar{x}$ ∈ Z_{5} is an example of group isomorphism.

**•** The map Φ: (Z, +) → (2Z, +) defined by Φ(n)=2n ∀ n ∈ Z is an isomorphism.

## Properties of Isomorphism

Property 1: If Φ: (G, 0) → (G′, *) is a group isomorphism, then the kernel of the map is trivial, that is, ker(Φ)={e_{G}}.

**Proof: **For a** **proof, visit the page: Injectivity criteria for homomorphism.

Property 2: If Φ: (G, 0) → (G′, *) is a group isomorphism, then we have:

- order of a = order of Φ(a) ∀ a ∈ G
- Both G and G′ have the same cardinality.

Property 3: Let Φ: (G, 0) → (G′, *) be a group isomorphism. Then the following are true:

- G is abelian if and only if G′ is abelian
- G is cyclic if and only if G′ is cyclic

**Remark:**

- We see that both abelian and cyclic properties are preserved by a group isomorphism.
- If a is a generator of G, then Φ(a) is a generator of G′.

Property 4: If Φ: (G, 0) → (G′, *) is a group isomorphism, then the inverse map Φ^{-1}: (G′, *) → (G, 0) is also an isomorphism.

Property 5: The composition of two isomorphisms is an isomorphism.

## Non Isomorphic Groups

**Example 1:** The groups (Z, +) and (Q, +) are not isomorphic.

**Solution:**

We know that (Z, +) is a cyclic group whereas (Q, +) is a non-cyclic group, see the page on cyclic groups. As the cyclic property is preserved by an isomorphism, we conclude that both the additive groups Z and Q are not isomorphic.

**Example 2:** The groups (Q, +) and (R, +) are not isomorphic.

**Solution:**

If there is an isomorphism between the additive groups Q and R, then they must have the same cardinality. But one knows that both Q and R have different cardinalities. So (Q, +) and (R, +) cannot be isomorphic.

**Example 3:** The groups (Q, +) and (Q^{+}, .) are not isomorphic.

**Solution:**

**Example 4:** The groups (R^{×}, .) and (R, +) are not isomorphic.

**Solution:**

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