Abstract Algebra/Group Theory/Homomorphism/Image of a Homomorphism is a Subgroup

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Theorem

Let f be a [[../Definition of Homomorphism, Kernel, and Image|homomorphism]] from [[../../Group/Definition of a Group|group]] G to [[../../Group/Definition of a Group|group]] K. Let eK be [[../../Group/Definition of a Group/Definition of Identity|identity]] of K.

[[../Definition of Homomorphism, Kernel, and Image|imf={kK|gG:f(g)=k}]] is a [[../../Subgroup/Definition of a Subgroup|subgroup]] of K.

Proof

Identity

0. f(eG)=eK [[../Homomorphism Maps Identity to Identity|homomorphism maps identity to identity]]
1. eK{kK|gG:f(g)=k} 0. and [[../../Group/Definition of a Group/Definition of Identity|eGG]]

2. Choose i{kK|gG:f(g)=k}||
3. iK
2.
4. eKi=ieK=i
[[../../Group/Definition of a Group/Definition of Identity#Usage3|i is in K and eK is identity of K(usage3)]]

5. iimf:eKi=ieK=i 2, 3, and 4.
6. eK is identity of imf [[../../Group/Definition of a Group/Definition of Identity#Usage4|definition of identity(usage 4)]]

Inverse

0. Choose i{kK|gG:f(g)=k}
1. gG:f(g)=i
0.
2. f(g)f(g1)=f(g1)f(g)=eK
[[../Homomorphism Maps Inverse to Inverse|homomorphism maps inverse to inverse]] between G and K
3. if(g1)=f(g1)i=eK
[[../Homomorphism Maps Inverse to Inverse|homomorphism maps inverse to inverse]]
4. i has inverse f( k-1) in im f
2, 3, and eK is identity of im f
5. Every element of im f has an inverse.

Closure

0. Choose i1,i2{kK|gG:f(g)=k}
1. g1,g2G:f(g1)=i1,f(g2)=i2
0.
2. g1g2G
Closure in G
3. f(g1g2){kK|gG:f(g)=k}
4. i1i2=f(g1)f(g2)=f(g1g2)
f is a homomorphism, 0.
5. i1i2imf
3. and 4.

Associativity

0. im f is a subset of K
1. is associative in K
2. is associative in im f 1 and 2

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