# 4.3 : Homomorphisms

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##### Definition: Homomorphism

Let $$(G_1,\star_1)$$, $$(G_2, \star_2)$$ be groups.

Then a function $$h:G_1 \rightarrow G_2$$ s.t. $$h(g_1)=g_2, g_1 \in G_1 \text{and } g_2 \in G_2$$ is called a homomorphism from the group $$G_1$$ to the group $$G_2$$ if $$h(g_1 \star_1g_1^{\shortmid})=h(g_1)\star_2 h(g_1^{\shortmid}), \;g_1,g_1^{\shortmid} \in G_1$$.

Further, if $$h$$ is a bijection and also a homomorphism, then $$h$$ is called an isomorphism.  In this case, we say $$G_1 \cong G_2$$, which means that $$G_1$$ is congruent to $$G_2$$.

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

Let $$C_n=<a,a^n=e>$$ be a cyclic group.

Define: $$h: C_n \rightarrow \mathbb{Z}_n$$ by $$h(a^k)=k \pmod{n}$$.  Then

1. Is $$h$$ a homomorphism?

2. Is $$h$$ injective (1-1)?

3. Is $$h$$ surjective (onto)?

If all three are true, then $$h$$ is an isomorphism.

Proof:

We will show a finite cyclic group is always isomorphic to $$(\mathbb{Z}_n, + \pmod{n})$$.

Proof of Homomorphism:

Let $$g_1,g_2 \in C_n$$.

We shall show that $$h(g_1\star_1 g_2)=h(g_1)\star_2h(g_2)$$.

Consider $$g_1=a^{k_1}$$ and $$g_2=a^{k_2}$$.

Then $$h(g_1)=k_1 \pmod{n}$$ and $$h(g_2)=k_2 \pmod{n}$$.

$$g_1\star_1 g_2=a^{k_1} \star_1 a^{k_2}=a^{k_1+k_2}$$.

Thus $$h(g_1\star_1g_2)=k_1+k_2 \pmod{n}$$.

Now consider $$h(g_1)\star_2 h(g_2)=k_1 \pmod{n} +k_2 \pmod{n}$$

$$=k_1+k_2 \pmod{n}$$.

Hence $$h(g_1\star_1 g_2)=h(g_1)\star_2h(g_2)$$.

Hence $$h$$ is a homomorphism.◻

Proof of Injection (1-1):

Let $$g_1, g_2 \in G$$ s.t. $$h(g_1)=h(g_2)$$.

Since $$g_1, g_2 \in C_n$$, $$g_1=a^{k_1}$$ and $$g_2=a^{k_2}, \; k_1,k_2 \in \mathbb{Z}$$.

Now, $$h(g_1)=k_1 \pmod{n}=k_2 \pmod{n}=h(g_2)$$.
Therefore, $$k_1=k_2 \in \mathbb{Z}_n$$.

Hence $$h$$ is injective (1-1).◻

Proof of Surjective (onto):

Let $$k \in \mathbb{Z}_n$$.

Since $$a^k \in C_n$$ by definition $$h(a^k)=k \pmod{n}$$.

Hence $$h$$ is surjective.◻

Since a finite cyclic group is homomorphic, injective and surjective to $$(\mathbb{Z}_n, + \pmod{n})$$, it is isomorphic to $$(\mathbb{Z}_n, + \pmod{n})$$. ◻

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

Define  $$h: (\mathbb{Z}, +) \to ( \mathbb{Z}_n, \cdot (mod n)$$ defined by

$$h(k)= a^k,$$ where $$\mathbb{Z}_n = <a>, k \in \mathbb{Z}.$$ Then

For $$k, m \in \mathbb{Z}, h(k+m)= a^{k+m}= a^k a^m= h(k)h(m).$$ Thus $$h$$ is a homomorphism.

Notice that

 $$0$$ $$\pm n$$ $$\pm 2n$$ $$\pm 3n$$ . . . . $$1$$ $$1 \pm n$$ $$1 \pm 2n$$ $$1 \pm 3n$$ . . . . $$\cdots$$ $$\cdots$$ $$n-1$$ $$n-1 \pm n$$ $$n-1 \pm 2n$$ $$n-1 \pm 3n$$ . . . . $$\mathbb{Z}$$ $$\downarrow$$ $$\downarrow$$ $$\downarrow$$ $$\downarrow$$ $$\downarrow$$ $$a^0$$ $$a^1$$ $$a^{n-1}$$ $$\mathbb{Z}_n$$

Hence $$h$$ is onto but not one to one.

Properties of homomorphisms:

##### Theorem $$\PageIndex{1}$$

Let $$h: G_1 \rightarrow G_2$$ be a homomorphism.  Then:

1. $$h(e_1)=e_2$$, where $$e_1, e_2$$ are identities of $$G_1, G_2$$ respectively.

2. $$h(g_1^{-1})=h(g_1)^{-1}, g_1 \in G_1$$.

3. $$h(g^n)=h(g)^n$$, $$n \in \mathbb{Z}$$

Proof:

1. Since $$e_1$$ is the identity of $$G_1$$, $$e_1 e_1=e_1.$$

Since $$h$$ is a homomorphism, $$h(e_1e_1)= h(e_1) h(e_1)=h(e_1)=e_2 h(e_1).$$

By cancellation law, $$e_2 = h(e_1).$$

1. Let $$g_1 \in G_1.$$ Then $$g_1g_1^{-1}=e_1.$$

Since $$h$$ is a homomorphism, $$h(g_1g_1^{-1})=h(g_1) h(g_1^{-1}) = h(e_1)=e_2.$$

Hence $$(h(g_1))^{-1} = h(g_1^{-1}).$$

1. Use induction on $$\mathbb{Z}_+$$.

##### Definition:

Let $$h: G_1 \to G_2$$ be a homomorphism. Then

1. The kernel of $$h$$ is defined as $$ker(h)=\{g_1 \in G_1| h(g_1)=e_2 \}.$$

2. The image  of $$h$$ is defined as $$im(h)= \{ g_2 \in G_2| h(g_1)=g_2, g_1 \in G_1\}.$$

##### Theorem $$\PageIndex{1}$$

Let $$h: G_1 \to G_2$$ be a homomorphism. Then $$ker(h) \leq G_1$$ and $$im(h) \leq G_2.$$ Also, $$\ker (h)$$ is a normal subgroup of $$G _1$$.

Proof:

Let $$h: G_1 \to G_2$$ be a homomorphism.

Since $$h(e_1)=e_2, e_1 \in ker(h).$$ Let $$g_1, g_1^{'} \in ker(h).$$  Then $$h(g_1)=e_2$$ and $$h( g_1^{'}) =e_2.$$

Similarly, we can show that $$im(h) \leq G_2.$$

We shall show that $$g_1^{-1} g_1^{'} \in ker(h).$$

Consider $$h(g_1^{-1} g_1^{'})= h(g_1^{-1})h( g_1^{'}) = h(g_1)^{-1} h( g_1^{'}) = e_2^{-1} e_2=e_2.$$ Hence  $$g_1^{-1} g_1^{'} \in ker(h).$$ Thus  $$ker(h) \leq G_1.$$

We shall show that $$\ker (h)$$ is a normal subgroup of $$G_1$$.

Let $$h \in \ker(h)$$ s.t. $$h(h)=e_H$$.

We will show $$ghg^{-1}\in \ker(h), \forall \; g \in G_1$$.

Consider $$h(ghg^{-1})=h(g) h(h) h(g^{-1})$$

$$= h(g) h(h) (h(g))^{-1}$$

$$=h(g) e_H (h(g))^{-1}$$

$$=h(g) (h(g))^{-1}$$

$$=e_H$$.

So $$\ker(h)$$ is a normal subgroup of $$G_1$$.

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