602: Range of Group Homomorphism Is Subgroup of Codomain|T.B.P.

Topics

About: group

The reader will have a description and a proof of the proposition that for any group homomorphism, the range of the homomorphism is a subgroup of the codomain.

G_{1}

\in {the groups}

G_{2}

\in {the groups}

f

: G_{1} \to G_{2}

\in {the group homomorphisms}

r a n f

= the range of f

//

Statements:

r a n f \in {the subgroups of G_{2}}

.
//

For any groups,

G_{1}, G_{2}

, and any group homomorphism,

f : G_{1} \to G_{2}

, the range of

f

r a n f

, is a subgroup of

G_{2}

The identity element of any group is denoted as

e

.

For any elements,

g_{1}, g_{2} \in r a n f

g_{1} g_{2} \in r a n f

, because

g_{j} = f (g_{j}^{'})

for a

g_{j}^{'} \in G_{1}

g_{1} g_{2} = f (g_{1}^{'}) f (g_{2}^{'}) = f (g_{1}^{'} g_{2}^{'})

e \in r a n f

, because

f (e) = e

.

For any element,

g \in r a n f

g^{- 1} \in r a n f

, because

g = f (g^{'})

for a

g^{'} \in G_{1}

g^{- 1} = {f (g^{'})}^{- 1} = f (g^{' - 1})

r a n f

satisfies the associative law, because the elements are elements of

G_{2}

and the operation is inherited from that of

G_{2}

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