# 4.5: Isormormophism theorems

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An isomorphism between two groups is a bijective mapping between two groups that preserves the group structure.

##### Definition: Term

Let $$G$$ and $$H$$ be groups. Let $$\phi: G \rightarrow H$$ be a group homomorphism.

An isomorphism from $$G$$ to $$H$$ is a function $$\phi: G \rightarrow H$$ that satisfies the following two properties:

1. \textit{Homomorphism property}: For any $$g_1, g_2 \in G$$, the operation in $$H$$ corresponding to $$\phi(g_1)$$ and $$\phi(g_2)$$ is the same as the operation in $$G$$ corresponding to $$g_1$$ and $$g_2$$. In other words, $$\phi(g_1g_2) = \phi(g_1) \cdot \phi(g_2)$$, where $$\cdot$$ represents the group operation.

2. \textit{Bijectivity property}: The function $$\phi$$ is one-to-one (injective) and onto (surjective), meaning that every element in $$H$$ is the image of exactly one element in $$G$$ under $$\phi$$.

If an isomorphism exists between two groups $$G$$ and $$H$$, we say that $$G$$ and $$H$$ are isomorphic, denoted by $$G \cong H$$. This means the two groups have the same group structure, even though their elements and operation symbols may differ.

Isomorphisms are significant in mathematics because they establish a correspondence between different groups, allowing us to understand the structure and properties of one group by studying the properties of another group.

Recall:  The kernel of $$\phi$$, denoted by $$\text{Ker}(\phi)$$, is defined as the set of elements in $$G$$ that map to the identity element of $$H$$. the image of $$\phi$$ is the set of all values that $\phi$ takes on as it maps elements from $$G$$ to $$H$$, denoted by $$\text{Im}(\phi) == \{\phi(g) ,|, g \in G\}$$.

##### Theorem $$\PageIndex{1}$$: First isomorphism Theorem

Let $$G$$ and $$H$$ be groups. Let $$\phi: G \rightarrow H$$ be a group homomorphism.  Then, the factor group $$G/\text{Ker}(\phi)$$ is isomorphic to the image of $$G,$$ under $$\phi$$, $$\text{Im}(\phi)$$. That is, $$G/\text{Ker}(\phi) \cong \text{Im}(\phi)$$

##### Example $$\PageIndex{1}$$

Consider the groups$$G = (\mathbb{Z}, +)$$ (the group of integers under addition) and$$H = (\mathbb{Z}/4\mathbb{Z}, +4)$$ (the group of integers modulo 4 under addition modulo 4). Define the homomorphism$$\varphi: \mathbb{Z} \rightarrow \mathbb{Z}/4\mathbb{Z}$$ as$$\varphi(x) = x \mod 4$$.

The kernel of$$\varphi$$ is$$\text{Ker}(\varphi) = {x \in \mathbb{Z} ,|, \varphi(x) = 0} = {x \in \mathbb{Z} ,|, x \mod 4 = 0} = 4\mathbb{Z}$$ (the subgroup of multiples of 4).

By applying the First Isomorphism Theorem, we have$$G/\text{Ker}(\varphi) \cong \text{Im}(\varphi)$$. In this case,$$G/\text{Ker}(\varphi)$$ is isomorphic to the image of$$G$$ under$$\varphi$$.

Using the theorem, we find that$$G/\text{Ker}(\varphi) \cong H$$, which means$$(\mathbb{Z}/4\mathbb{Z}, +4)$$ is isomorphic to itself.

##### Theorem $$\PageIndex{2}$$: Second isomorphism Theorem

Let $$G$$ be a group,$$H$$ and$$K$$ be subgroups of $$G$$ such that $$H \leq G$$. Then, the set $$HK = {hk ,|, h \in H, k \in K}$$ is a subgroup of$$G$$, and$$H \cap K$$ is a normal subgroup of$$H$$. Moreover, the factor group$$(HK)/(H \cap K)$$ is isomorphic to$$K/(H \cap K)$$.

That is, $$(HK)/(H \cap K) \cong K/(H \cap K)$$

##### Example $$\PageIndex{2}$$

Consider the group $$G = (\mathbb{Z}, +)$$ (the group of integers under addition). Let $$H = (2\mathbb{Z}, +)$$ be the subgroup of even integers, and $$K = (3\mathbb{Z}, +)$$ be the subgroup of multiples of 3.

In this case, $$H$$ and $$K$$ are both normal subgroups of $$G$$.

By applying the Second Isomorphism Theorem, we have $$(HK)/(H \cap K) \cong K/(H \cap K)$$. Here, $$(HK)/(H \cap K)$$ is isomorphic to $$K/(H \cap K)$$.

Using the theorem, we find that $$(2\mathbb{Z} + 3\mathbb{Z}) / (2\mathbb{Z} \cap 3\mathbb{Z}) \cong (3\mathbb{Z}) / (2\mathbb{Z} \cap 3\mathbb{Z)}$$.

In this example, $$(2\mathbb{Z} + 3\mathbb{Z}) / (2\mathbb{Z} \cap 3\mathbb{Z})$$ represents the set of all integers that are divisible by both 2 and 3, which is equivalent to $$(6\mathbb{Z}) / (6\mathbb{Z}) = {0}$$ (the trivial group). Similarly, $$(3\mathbb{Z}) / (2\mathbb{Z} \cap 3\mathbb{Z})$$ represents the set of all integers that are divisible by 3 but not divisible by 2, which is isomorphic to $$(\mathbb{Z}/3\mathbb{Z}, +3)$$ (the group of integers modulo 3 under addition modulo 3).

Therefore, we conclude that $${0} \cong \mathbb{Z}/3\mathbb{Z}$$.

##### Theorem $$\PageIndex{3}$$: Third isomorphism Theorem

Let $$G$$ be a group, and let $$N$$ and $$M$$ be normal subgroups of $$G$$, with $$N \leq M$$. Then, the factor group $$M/N$$ is isomorphic to a subgroup of $$G/N$$.

That is, $$M/N \cong (M/N) / (N/N)$$

##### Example $$\PageIndex{3}$$

Consider the group $$G = (\mathbb{Z}, +)$$ (the group of integers under addition). Let $$N = (6\mathbb{Z}, +)$$ be the normal subgroup of multiples of 6, and $$M = (9\mathbb{Z}, +)$$ be the normal subgroup of multiples of 9.

In this case, $$N \leq M$$.

By applying the Third Isomorphism Theorem, we have $$M/N \cong (M/N) / (N/N)$$.

Using the theorem, we find that $$(9\mathbb{Z}) / (6\mathbb{Z}) \cong (9\mathbb{Z} / 6\mathbb{Z}) / (6\mathbb{Z} / 6\mathbb{Z})$$.

The factor group $$(9\mathbb{Z}) / (6\mathbb{Z})$$ represents the set of all integers that are multiples of 9, modulo 6. This is isomorphic to $$(\mathbb{Z}/6\mathbb{Z}, +6)$$ (the group of integers modulo 6 under addition modulo 6).

Similarly, $$(9\mathbb{Z} / 6\mathbb{Z}) / (6\mathbb{Z} / 6\mathbb{Z})$$ represents the set of cosets of $$(6\mathbb{Z} / 6\mathbb{Z})$$ within $$(9\mathbb{Z} / 6\mathbb{Z})$$. Since $$(6\mathbb{Z} / 6\mathbb{Z})$$ is the trivial group, there is only one coset, which is $${6\mathbb{Z}}$$.

Therefore, we conclude that $$\mathbb{Z}/6\mathbb{Z} \cong {6\mathbb{Z}}$$ (the trivial group).

Ex

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