AATA Group Homomorphisms

Section 11.1 Group Homomorphisms

A homomorphism between groups

(G, \cdot)

and

(H, \circ)

is a map

ϕ : G \to H

such that

ϕ (g_{1} \cdot g_{2}) = ϕ (g_{1}) \circ ϕ (g_{2})

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for

g_{1}, g_{2} \in G .

The range of

ϕ

H

is called the homomorphic image of

ϕ .

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Two groups are related in the strongest possible way if they are isomorphic; however, a weaker relationship may exist between two groups. For example, the symmetric group

S_{n}

and the group

Z_{2}

are related by the fact that

S_{n}

can be divided into even and odd permutations that exhibit a group structure like that

Z_{2},

as shown in the following multiplication table.

\begin{array}{ccc} even & odd \\ even & even & odd \\ odd & odd & even \end{array}

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We use homomorphisms to study relationships such as the one we have just described.

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Example 11.1.

Let $G$ be a group and $g \in G .$ Define a map $ϕ : Z \to G$ by $ϕ (n) = g^{n} .$ Then $ϕ$ is a group homomorphism, since

ϕ (m + n) = g^{m + n} = g^{m} g^{n} = ϕ (m) ϕ (n) .

This homomorphism maps $Z$ onto the cyclic subgroup of $G$ generated by $g .$

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Example 11.2.

Let $G = G L_{2} (R) .$ If

A = (\begin{matrix} a & b \\ c & d \end{matrix})

is in $G,$ then the determinant is nonzero; that is, $det (A) = a d - b c \neq 0 .$ Also, for any two elements $A$ and $B$ in $G,$ $det (A B) = det (A) det (B) .$ Using the determinant, we can define a homomorphism $ϕ : G L_{2} (R) \to R^{*}$ by $A \mapsto det (A) .$

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Example 11.3.

Recall that the circle group $T$ consists of all complex numbers $z$ such that $| z | = 1 .$ We can define a homomorphism $ϕ$ from the additive group of real numbers $R$ to $T$ by $ϕ : θ \mapsto \cos θ + i \sin θ .$ Indeed,

\begin{aligned} ϕ (α + β) & = \cos (α + β) + i \sin (α + β) \\ = (\cos α \cos β - \sin α \sin β) + i (\sin α \cos β + \cos α \sin β) \\ = (\cos α + i \sin α) (\cos β + i \sin β) \\ = ϕ (α) ϕ (β) . \end{aligned}

Geometrically, we are simply wrapping the real line around the circle in a group-theoretic fashion.

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The following proposition lists some basic properties of group homomorphisms.

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Proposition 11.4.

Let $ϕ : G_{1} \to G_{2}$ be a homomorphism of groups. Then

If $e$ is the identity of $G_{1},$ then $ϕ (e)$ is the identity of $G_{2};$
For any element $g \in G_{1},$ $ϕ (g^{- 1}) = [ϕ (g)]^{- 1};$
If $H_{1}$ is a subgroup of $G_{1},$ then $ϕ (H_{1})$ is a subgroup of $G_{2};$
If $H_{2}$ is a subgroup of $G_{2},$ then $ϕ^{- 1} (H_{2}) = {g \in G_{1} : ϕ (g) \in H_{2}}$ is a subgroup of $G_{1} .$ Furthermore, if $H_{2}$ is normal in $G_{2},$ then $ϕ^{- 1} (H_{2})$ is normal in $G_{1} .$

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Proof.

(1) Suppose that $e$ and $e^{'}$ are the identities of $G_{1}$ and $G_{2},$ respectively; then

e^{'} ϕ (e) = ϕ (e) = ϕ (e e) = ϕ (e) ϕ (e) .

By cancellation, $ϕ (e) = e^{'} .$

(2) This statement follows from the fact that

ϕ (g^{- 1}) ϕ (g) = ϕ (g^{- 1} g) = ϕ (e) = e^{'} .

(3) The set $ϕ (H_{1})$ is nonempty since the identity of $G_{2}$ is in $ϕ (H_{1}) .$ Suppose that $H_{1}$ is a subgroup of $G_{1}$ and let $x$ and $y$ be in $ϕ (H_{1}) .$ There exist elements $a, b \in H_{1}$ such that $ϕ (a) = x$ and $ϕ (b) = y .$ Since

x y^{- 1} = ϕ (a) [ϕ (b)]^{- 1} = ϕ (a b^{- 1}) \in ϕ (H_{1}),

$ϕ (H_{1})$ is a subgroup of $G_{2}$ by Proposition 3.31.

(4) Let $H_{2}$ be a subgroup of $G_{2}$ and define $H_{1}$ to be $ϕ^{- 1} (H_{2});$ that is, $H_{1}$ is the set of all $g \in G_{1}$ such that $ϕ (g) \in H_{2} .$ The identity is in $H_{1}$ since $ϕ (e) = e^{'} .$ If $a$ and $b$ are in $H_{1},$ then $ϕ (a b^{- 1}) = ϕ (a) [ϕ (b)]^{- 1}$ is in $H_{2}$ since $H_{2}$ is a subgroup of $G_{2} .$ Therefore, $a b^{- 1} \in H_{1}$ and $H_{1}$ is a subgroup of $G_{1} .$ If $H_{2}$ is normal in $G_{2},$ we must show that $g^{- 1} h g \in H_{1}$ for $h \in H_{1}$ and $g \in G_{1} .$ But

ϕ (g^{- 1} h g) = [ϕ (g)]^{- 1} ϕ (h) ϕ (g) \in H_{2},

since $H_{2}$ is a normal subgroup of $G_{2} .$ Therefore, $g^{- 1} h g \in H_{1} .$

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Let

ϕ : G \to H

be a group homomorphism and suppose that

e

is the identity of

H .

By Proposition 11.4,

ϕ^{- 1} ({e})

is a subgroup of

G .

This subgroup is called the kernel of

ϕ

and will be denoted by

\ker ϕ .

In fact, this subgroup is a normal subgroup of

G

since the trivial subgroup is normal in

H .

We state this result in the following theorem, which says that with every homomorphism of groups we can naturally associate a normal subgroup.

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Theorem 11.5.

Let $ϕ : G \to H$ be a group homomorphism. Then the kernel of $ϕ$ is a normal subgroup of $G .$

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Example 11.6.

Let us examine the homomorphism $ϕ : G L_{2} (R) \to R^{*}$ defined by $A \mapsto det (A) .$ Since $1$ is the identity of $R^{*},$ the kernel of this homomorphism is all $2 \times 2$ matrices having determinant one. That is, $\ker ϕ = S L_{2} (R) .$

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Example 11.7.

The kernel of the group homomorphism $ϕ : R \to C^{*}$ defined by $ϕ (θ) = \cos θ + i \sin θ$ is ${2 π n : n \in Z} .$ Notice that $\ker ϕ ≅ Z .$

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Example 11.8.

Suppose that we wish to determine all possible homomorphisms $ϕ$ from $Z_{7}$ to $Z_{12} .$ Since the kernel of $ϕ$ must be a subgroup of $Z_{7},$ there are only two possible kernels, ${0}$ and all of $Z_{7} .$ The image of a subgroup of $Z_{7}$ must be a subgroup of $Z_{12} .$ Hence, there is no injective homomorphism; otherwise, $Z_{12}$ would have a subgroup of order $7,$ which is impossible. Consequently, the only possible homomorphism from $Z_{7}$ to $Z_{12}$ is the one mapping all elements to zero.

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Example 11.9.

Let $G$ be a group. Suppose that $g \in G$ and $ϕ$ is the homomorphism from $Z$ to $G$ given by $ϕ (n) = g^{n} .$ If the order of $g$ is infinite, then the kernel of this homomorphism is ${0}$ since $ϕ$ maps $Z$ onto the cyclic subgroup of $G$ generated by $g .$ However, if the order of $g$ is finite, say $n,$ then the kernel of $ϕ$ is $n Z .$

Abstract Algebra: Theory and Applications

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