4.3: Homomorphisms

Definition: Homomorphism

Let \((G_1,\star_1)\), \((G_2, \star_2)\) be groups. Then a function \(h: G_1 \rightarrow G_2\) such that \(h(g_1)=g_2, g_1 \in G\) and \( g_2 \in G_2\) is called a homomorphism from the group \(G_1\) to the group \(G_2\) if 

\(h(g_1 \star_1g_1^{\shortmid})=h(g_1)\star_2 h(g_1^{\shortmid}), \) for \(\;g_1,g_1^{\shortmid} \in G_1\).

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

Moreover, if \(h\) is an isomorphism from a group \(G\) to itself, then \(h\) is called an automorphism. \(Aut(G)=\) Set of all automorphisms=\(\{h: G \rightarrow G \mid h \text{ is an isomorphism}\}.\)

 

Example \(\PageIndex{1}\)

Let \(C_n=\{\langle a\rangle \mid a^n=e\}\). 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.

Answer

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  \( h: (\mathbb{Z}, +) \to ( \mathbb{Z}_n, \cdot \pmod{n} )\) 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 

 

Table \(\PageIndex{2}\): Homomorphism

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

 

Hence \( h \) is onto but not one to one.

Note

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

Then a function \(h: G_1 \rightarrow G_2\) s.t. \(h(g_1)=e_2, \forall g_1 \in G_1 \text{and  the identity } e_2 \in G_2\) is called a trivial  homomorphism from the group \(G_1\) to the group \(G_2\).

Example \(\PageIndex{3}\)

Let \(G\) be a group. The mapping \(1_G: G \to G\) defined by \(1_G(g)=g, \forall g \in G\) is a homomorphism.

Example \(\PageIndex{4}\)

Define a mapping from \(h: D_4 \rightarrow D_4\) by \(h(a)=a^3, a \in D_4\). Is \(h\) a homomorphism?

Solution

No.   \(D_4=\{ e, r,r^2, r^3, s, sr, sr^2, sr^3\}\) such that \(r^4=e, s^2=e, srs=r^{-1}\). Consider \(h(s)=s^3=s, h(r)=r^3\)   and \(h(sr)=(sr)^3=(sr)(sr)(sr)= (srs)r(sr)=r^2r(sr)=r^3(r^2s)=rs=sr^2\). But \(h(s)h(r)=sr^3\).

Example \(\PageIndex{5}\)

Define a mapping from \(h: S_4 \rightarrow S_4\) by \(h(\sigma)=\sigma^2\), for \(\sigma \in S_4\). Is \(h\) a homomorphism?

Solution

No.  \(h((1, 2))=h((2, 3))=e\) but \(h((1, 2)(2, 3))=h((1, 2, 3))=(3, 2, 1).\)

Example \(\PageIndex{6}\)

Let \(\phi: \mathbb{Z}_5 \rightarrow \mathbb{Z}_2\) mapping defined by \(\phi(a)=\left\{ \begin{array}{cc} 1 & if \, a \,\mbox{is odd}\\0 & if \, a \, \mbox{is even}\end{array}\right. \).

Show that \(\phi\) is not a homomorphism.

Solution

\(\phi(2+4)=\phi(1)=1\) but \(\phi(2)+\phi(4)=0.\)

Properties of homomorphisms:

Theorem \(\PageIndex{1}\)

Let \((G_1,\star_1)\), \((G_2, \star_2)\) be groups and 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. \(h(g_1^n)=h(g_1)^n\), \( n \in \mathbb{Z} \) and  \(g_1 \in G_1\).

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, \(e_2 = h(e_1). \)

2. 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 \( (h(g_1))^{-1} = h(g_1^{-1}). \)

3. Use induction on \( \mathbb{Z}_+ \).

 

Corollary\(\PageIndex{1}\)

Let \((G_1,\star_1)\), \((G_2, \star_2)\) be groups and let \(h: G_1 \rightarrow G_2\) be a homomorphism.  If \(g_1 \in G_1 \) has a finite order then \(h(g_1)\) has a finite order, and \(|h(g_1)|\) divides \(|g_1|\).

Proof:

Let \((G_1,\star_1)\), \((G_2, \star_2)\) be groups and let \(h: G_1 \rightarrow G_2\) be a homomorphism. Assume that \(g_1 \in G_1 \) has a finite order \(n.\) Then  \(g_1^n=e_1.\)  Hence \(h(g_1)^n=h(g_1^n) =h(e_1)=e_2.\)  By Theorem 2.4.5(2.4: Cyclic groups), \(|h(g_1)|\) divides \(n\). Hence \(|h(g_1)|\) divides \(|g_1|\).

The following theorem allows us to determine if two homomorphisms are equal by checking only the generators instead of every element in the domain.

Note

 Recall: Let \((G,\star)\) be a group.

Let \(X=\{x_1, x_2, \cdots, x_n\}\) where \(x_1, x_2 \cdots, x_n \in G\), then the subset  of \(G\) generated by \(X\) denoted by \(\langle X \rangle \) is

\( \left\{x_1^{m_1} x_2^{m_2} \cdots x_n^{m_n} \mid m_1, m_2 \cdots, m_n \in \mathbb{Z} \right\}.\)

Theorem \(\PageIndex{2}\)

Let  \(h: G\to G_1\) and  \(h_1: G\to G_1\) be  group homomorphisms and \(G=\langle X\rangle\) is generated by a subset \(X\), then  show that \(h=h_1\) if and only if   \(h(x)=h_1(x), \forall x \in X.\)

Proof:

Let \(X=\{g_1, \cdots,  g_n  \}.\) Assume \(h(g_i) = h_1(g_i), \forall g_i \in X, i \in \mathbb{Z} \). Let \(g \in G\). Since \(G = \langle X \rangle, g= g_1^{m_1}\cdots g_n^{m_n} ,\) where \(m_1,\dots,m_n \in \mathbb{Z}\). Consider \(h(g) = h(g_1^{m_1} \dots g_n^{m_n}) = h(g_1^{m_1}) \cdots h(g_n^{m_n}), \) as h is a homomorphism. Then;\begin{align*}h(g) &= (h(g_1))^{m_1}\dots (h(g_n))^{m_n} \\&= ((h_1(g_1))^{m_1} \cdots (h_1(g_n))^{m_n} &\text{(by assumption)}\\&= h_1(g_1^{m_1}) \cdots h_1(g_n^{m_n}),\\&= h_1(g_1^{m_1} \dots g_n^{m_n})\\&= h_1(g).\end{align*} Hence, \( h(g) = h_1(g), \forall g \in G.\) The converse is clearly true. Hence the result.

Example \(\PageIndex{7}\)

Describe all group homomorphisms from \(\mathbb{Z}\) to itself.  From which, how many of them are surjective?

Solution

Since \(\mathbb{Z}= \langle 1 \rangle,\) any such group homomorphism \(h\) determined by \(h(1).\)  If  \(h(1)=k, k \in \mathbb{Z}\), then \(h(n)=kn, \forall n \in \mathbb{Z}.\) Hence there is such a group homomorphism for every \(k \in \mathbb{Z}\).  The only surjective group homomorphisms are \(k=\pm 1\). 

Example \(\PageIndex{8}\)

Find all group homomorphisms from \(C_3\) to \(A_4\), where \(C_3=\{\langle a\rangle \mid a^3=e\}\).

Solution

 Let \(C_3= \langle c \rangle\). Any group homomorphism \(h:C_3 \to G\), any group \(G\), determined by \(h(c),\)  and \(h(c)\) must be an element of order \(1\) or \(3\) in \(G\). If \(h(c)=e, e \in A_4\), then \(h\) is a trivial homomorphism. Otherwise, \(h(c)\) must be a \(3-\) cycle in \(A_4\). There are \(\dfrac{4!}{3^11^11!1!}=8\), \(3-\)cycle permutations in \(A_4\). Hence, eight such group homomorphisms.

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 \(\PageIndex{2}\)

 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 \(x \in \ker(h) \) s.t. \(h(x)=e_2\).

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

Consider \(h(gxg^{-1})=h(g) h(x) h(g^{-1}) \)

                               \(= h(g) h(x) (h(g))^{-1} \)

                               \(=h(g) e_2 (h(g))^{-1} \)

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

                               \(=e_2 \).

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

 

Example \(\PageIndex{9}\)

 Let \( h: G_1 \to G_2\) be a group homomorphism with \(K= ker(h). \) Show that for any \(g_1 \in  G_1\), \(g_1K=\{g \in G_1 \mid h(g)=h(g_1)\}.\)

Solution

Let \(g_1 \in  G_1\). Let \(g \in  g_1 K\). Then \(g=g_1k\) where \(k\in ker(h).\) Thus, \(h(k)=e_2.\) Consider \(h(g)= h(g_1k)=h(g_1) h(k)= h(g_1) e_2=h(g_1).\)  Hence, \(g_1K \subset \{g \in G_1 \mid h(g)=h(g_1)\}.\) Conversely, suppose  \( x \in \{g \in G_1 \mid h(g)=h(g_1)\}.\) Then \(h(x)= h(g_1) \implies h(g_1^{-1}x)=h(g_1^{-1})h(x)= \left(h(g_1)\right)^{-1}h(x)=e_2.\) Hence, \(g_1^{-1} x\in K\). Thus \(x \in g_1 K.\) Hence the reuslt.

Theorem \(\PageIndex{3}\)

Let \( h: G_1 \to G_2\) be a homomorphism. Then \( ker(h)=\{e_1\}\) if and only if \(h\) is injective.

Proof

Let \( h: G_1 \to G_2\) be a homomorphism. Assume that \( ker(h)=\{e_1\}\). We shall show that \(h\) is injective. Let \(x,y\in G_1\) such that \(h(x)=h(y).\) Then \(h(xy^{-1)= e_2.\) Thus, \(xy^{-1) \in ker(h)\). Hence \(xy^{-1)=e_1 \implies x=y.\) Thus \(h\) is injective. Conversely, assume that \(h\) is injective. Since \(h(e_1=e_2\),  \( ker(h)=\{e_1\}\). Hence the result.

Note

If both \(G_1\) and \(G_2\) are finite and \(h\) is an injective homomorphism (is called monomorphism) then \(h\) is an isomorphism.

Example \(\PageIndex{10}\)

Find all group homomorphisms from \(A_4\) to \(C_3\), where \(C_3=\{\langle a\rangle \mid a^3=e\}\).

Solution

The kernel of a group homomorphism \(h:A_4 \to G\), any group \(G\), is a normal subgroup of \(A_4\). Hence \(ker(h)\) is either \(\{e\}, K=\{e,(12)(34),(13)(24),(14)(23)\}\) or \(A_4\). Since \(|A_4| \ne |C_3|, \)  \( ker(h) \ne \{e\}\). Suppose  \(ker(h)=K\). Let \(\phi: A_4 \to A_4/K\) be the coset map defined by \(\phi(\sigma)= \sigma K,\) for \( \sigma \in A_4.\)  Note that \( |A_4/K|=3,\) and \(A_4/K=\{K, (1\,2 \, 3)K, (1\, 2\, 3)^2K\}=\langle (1\,2 \, 3)K \rangle \). There are exactly two isomorphisms from \(A_4/K\) to \(C_3\) in the following way:

Let \(\bar{\phi}: A_4 /K \to C_3\) be the isomorphism such that \(h= \bar{\phi} \circ \phi\), then \(\bar{\phi}\left( (1\,2 \, 3)K\right)\) is either \(a\) or \(a^2\) giving two group homomorphisms \[h_1(x)=\left\{\begin{array}{ll} e &\mbox{ if } x\in K\\ a  &\mbox{ if } x\in (1\,2 \, 3)K\\ a^2 & \mbox{ if } x\in  (3\,2 \, 1)K\\ \end{array} \right.\], and \[h_1(x)=\left\{\begin{array}{ll} e &\mbox{ if } x\in K\\ a^2  &\mbox{ if } x\in (1\,2 \, 3)K\\ a & \mbox{ if } x\in  (3\,2 \, 1)K\\ \end{array} \right.\].

If \(ker(h)= A_4\), then \(h\) is a trivial homomorphism.

Isomorphic Groups

As we said earlier in Chapter 2, we can use the Cayley tables to identify isomorphic groups. For example:

Cayley table for Klein \(4\)-group: \(H=\left\{\langle a,b, c \rangle \mid a^2=e, b^2=e, c^2=e \right.\) and \( \left.ab=ba=c, bc=cb=a, ac=ca=b\right\} =\{e, a, b, c\}.\)

Table 1: Cayley table for Klein \(4\)-group
	\(e\)	\(a\)	\(b\)	\(c\)
\(e\)	\(e\)	\(a\)	\(b\)	\(c\)
\(a\)	\(a\)	\(e\)	\(c\)	\(b\)
\(b\)	\(b\)	\(c\)	\(e\)	\(a\)
\(c\)	\(c\)	\(b\)	\(a\)	\(e\)

Compare it with the Cayley tables for  \((\mathbb{Z}_4, +(mod\, 4))\) and \((U(5),\cdot (mod 5)\).

Table 2:  Cayley table for  \((\mathbb{Z}_4, +(mod\, 4))\)
\(+(mod\, 4)\)	\(0\)	\(1\)	\(2\)	\(3\)
\(0\)	\(0\)	\(1\)	\(2\)	\(0\)
\(1\)	\(1\)	\(2\)	\(3\)	\(0\)
\(2\)	\(2\)	\(3\)	\(0\)	\(1\)
\(3\)	\(3\)	\(0\)	\(1\)	\(2\)

\(U(5)=\{1, 2, 3, 4\}\).

Table 3:  Cayley table for  \((U(5), \cdot(mod\, 4))\)
\(\cdot (mod\, 5)\)	\(1\)	\(2\)	\(3\)	\(4\)
\(1\)	\(1\)	\(2\)	\(3\)	\(4\)
\(2\)	\(2\)	\(4\)	\(1\)	\(3\)
\(3\)	\(3\)	\(1\)	\(4\)	\(2\)
\(4\)	\(4\)	\(3\)	\(2\)	\(1\)

We can see that Tables 2 and 3 are identical (mapping: \(0 \mapsto 1,1 \mapsto 2, 2 \mapsto 4, 3  \mapsto 3)\), while Table 1 is different. Hence, we say that the groups \(\mathbb{Z}_4\) and \(U(5)\) are isomorphic and denoted by  \(\mathbb{Z}_4 \cong U(5)\), and the groups \(\mathbb{Z}_4\) and \(K_4\) are not isomorphic and denoted by  \(\mathbb{Z}_4 \not\cong K_4\).

Definition:  Isomorphic Groups

Two groups \(G\) and \(H\) are isomorphic, denoted by \(G\cong H\) ,if there exists an isomorphism from \(G\) to \(H\).

Example \(\PageIndex{11}\)

\(\mathbb{Z}_4 \cong U(5)\)

Example \(\PageIndex{12}\)

\(\mathbb{Z}\cong 4\mathbb{Z}\)

Answer

The function \(\phi: \mathbb{Z} \to 4\mathbb{Z}\) , \(\phi(x)=4x, x \in \mathbb{Z}\) is clearly a bijection. 

Theorem \(\PageIndex{4}\)

Let \(G\) be a group, and let \(g \in G\). Then

1. If \(|g|\) is infinite then \(\langle g\rangle \cong \mathbb{Z}.\)

Proof

Use \(\phi: \langle g\rangle \to \mathbb{Z}\) such that \(\phi(g^k)=k.\) 

2. If \(|g|=n\) then  \(\langle g\rangle \cong \mathbb{Z}_n.\)

Proof

Use \(\phi: \langle g\rangle \to \mathbb{Z}_n\) such that \(\phi(g^k)=k (mod \, n).\) 

Theorem \(\PageIndex{5}\)

1. Let \(G\) be group of order \(4\) then either \(G\) is cyclic or \(G \cong K_4.\) 

2. Let \(G\) be group of order \(6\) then either \(G\) is cyclic or \(G \cong D_3.\) 

 Examples of structural properties of a group \(G\):

1. \(G\) is abelian.
2. \(G\) is cyclic.
3. \(G\) is finite.
4. \(|G|=n\).  
5. \(G\) has no element of order \(k\).

and many more.......

Theorem \(\PageIndex{6}\)

If two groups \(G\) and \(H\) are isomorphic and \(G\) has a structural property, then \(H\) has the structural property.

Theorem \(\PageIndex{7}\): Cayley's Theorem

Every group of order \(n\) is isomorphic to a subgroup of \(S_n.\)

Example \(\PageIndex{13}\)

Find all isomorphisms from \(\mathbb{Z}\) to \(U(7)\).

Solution

Add example text here.

Example \(\PageIndex{14}\)

Prove or disprove: \(S_4\) and \(D_{12}\) are isomorphic.

Solution

Even though \(|S_4|=|D_{12}|=24\) and both are non abelian, \(S_4 \not\cong D_{12}\). Since \(D_{12}=\{\langle r,s\rangle \mid r^{12}=e, s^2=e, srs=r^{-1}\}, r \in D_{12}\)  and \(|r|=12.\)

Now, we shall show that the order of all the elements of \(S_4\) is less than or equal to \(4\). This can be achieved by examining all possible types of permutations and their corresponding orders. The partitions of 4 are \(4, 3+1, 2+2, 2+1+1\) and \( 1+1+1+1.\) Hence the possible orders of permutations are \(4, lcm(3, 1), lcm(2, 2), lcm(2, 1, 1)\) and \(lcm(1, 1, 1,1).\) Thus the possible orders of permutations are \(4, 3, 2, 1.\) Hence the order of all the elements of \(S_4\) is less than or equal to \(4\). Therefore, \(S_4\) doesn't have an element of order \(12.\)  Hence, \(S_4\) and \(D_{12}\) are not isomorphic.

Example \(\PageIndex{15}\)

Prove or disprove: \((\mathbb{Q}, +)\) and \((\mathbb{Q}\setminus\{0\}, \cdot)\) are isomorphic.

Solution

Let \(\phi: (\mathbb{Q}, +) \rightarrow (\mathbb{Q}\setminus\{0\}, \cdot)\) be an isomorphism. Since \(2 \in \mathbb{Q}\setminus\{0\}\) and \(\phi\) is subjective, there exists \(a \in \mathbb{Q}\) such that \(\phi(a)=2.\) Now, \(\phi(a)=\phi\left(\dfrac{a}{2}+ \dfrac{a}{2}\right)= \phi \left(\dfrac{a}{2}\right)\phi\left(\dfrac{a}{2}\right).\) Thus, \(\phi\left(\dfrac{a}{2}\right)^2=2.\)  Hence \(\phi \left(\dfrac{a}{2}\right)=\sqrt{2}\not\in \mathbb{Q}.\) This is a contradiction.  Hence  \((\mathbb{Q}, +)\) and \((\mathbb{Q}\setminus\{0\}, \cdot)\) are not isomorphic.

Automorphisms

Example \(\PageIndex{16}\)

Let \(G\) be a finite abelian group of order \(n\). Let \(\phi_m: G \rightarrow G\) be the \(m\)-power map \(\phi_m(g)=g^m\). Then 

1. Show that \(\phi_m\) is a homomorphism.
2. Show that \(ker(\phi_m)=\{g \in G \mid g^d=e, d=gcd(m.n)\}.\)
3. If \(m\) and \(n\) are relatively prime, then show that \(\phi_m\) is an automorphism.

 

Definition: 

\(Aut(G)\) is defined as the set of all automorphisms of a group \(G\).

Note

The identity homomorphism \(1_G\ \in Aut(G).\)

Theorem \(\PageIndex{8}\)

Let \(G\) be a group. Then \(Aut(G)\) forms a group under the composition of functions.

Example \(\PageIndex{17}\)

Let \(G\) be a group and let \(a \in G\) then show that \(\sigma_a :G \to G\) defined by \(\sigma_a(g)=aga^{-1}\)is an isormorphism.

Solution

Let \(g, g'\in G\), consider

\begin{align*}\sigma_a(g) \sigma_a(g') &= aga^{-1}ag'a^{-1} \\&= ageg'a^{-1} \\&= a(gg')a^{-1} \\&= \sigma_a(gg')\end{align*}

Thus, \(\sigma_a\) is a homomorphism. Now we will show that \(\sigma_a\) is bijective. Let \(g, g' \in G\) such that \(\sigma_a(g) = \sigma_a(g')\). We will show that \(\sigma_{a}(g) = \sigma_{a}(g') \implies g = g'. \)

\begin{align*} \sigma_a(g) &= \sigma_a(g') \\ aga^{-1} &= ag'a^{-1} \\ a^{-1}aga^{-1}a &= a^{-1}ag'a^{-1}a \\ege &= eg'e \\ g &= g'\end{align*}

Hence, \(\sigma_a\) is injective. Now, let \(h \in G\), we will show that \(\exists \ g \in G\) such that \(\sigma_a(g) = h\). Let \(g = a^{-1}ha\), then

\begin{align*}\sigma_a(g) &= aga^{-1} \\&= a(a^{-1}ha)a^{-1} \\&= ehe \\&= h\end{align*}

Thus, \(\sigma_a\) is a surjective homomorphism. Hence \(\sigma_a: G \mapsto G\) is an isomorphism. 

 

Definition: 

An inner automorphism of a group \(G\) is defined as, \(h_a: G \to G, a \in G \), \(h_a(x)=axa^{-1}, \forall x \in G.\)

\(Inn(G)=\{h_a: G \to G \mid h_a(x)=axa^{-1}, a,x \in G\},\) is the set of all inner automorphisms of a group \(G\)

Theorem \(\PageIndex{9}\)

Let \(G\) be a group. Then \( Inn(G) \le Aut(G)\).