Abstract Algebra/Group Theory/Homomorphism/Image of a Homomorphism is a Subgroup
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|]] is a [[../../Subgroup/Definition of a Subgroup|subgroup]] of K.
Proof
Identity
0. [[../Homomorphism Maps Identity to Identity|homomorphism maps identity to identity]] 1. 0. and [[../../Group/Definition of a Group/Definition of Identity|]] - 2. Choose ||
- 3.
2. - 4.
[[../../Group/Definition of a Group/Definition of Identity#Usage3|i is in K and eK is identity of K(usage3)]] 5. 2, 3, and 4. 6. is identity of [[../../Group/Definition of a Group/Definition of Identity#Usage4|definition of identity(usage 4)]]
Inverse
0. Choose - 1.
0. - 2.
[[../Homomorphism Maps Inverse to Inverse|homomorphism maps inverse to inverse]] between G and K - 3.
[[../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 - 1.
0. - 2.
Closure in G - 3.
- 4.
f is a homomorphism, 0. - 5.
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