Editing Kernel (algebra) (section)

=== Group homomorphisms ===
{{Group theory sidebar}}
Let ''G'' and ''H'' be [[group (mathematics)|group]]s and let ''f'' be a [[group homomorphism]] from ''G'' to ''H''. If ''e''<sub>''H''</sub> is the [[identity element]] of ''H'', then the ''kernel'' of ''f'' is the preimage of the singleton set {''e''<sub>''H''</sub>}; that is, the subset of ''G'' consisting of all those elements of ''G'' that are mapped by ''f'' to the element ''e''<sub>''H''</sub>.<ref name=":0">{{Cite book |last1=Dummit |first1=David Steven |title=Abstract algebra |last2=Foote |first2=Richard M. |date=2004 |publisher=Wiley |isbn=978-0-471-43334-7 |edition=3rd |location=Hoboken, NJ}}</ref><ref name=":1">{{Cite book |last1=Fraleigh |first1=John B. |title=A first course in abstract algebra |last2=Katz |first2=Victor |date=2003 |publisher=Addison-Wesley |isbn=978-0-201-76390-4 |edition=7th |series=World student series |location=Boston}}</ref><ref name=":2">{{Cite book |last=Hungerford |first=Thomas W. |title=Abstract Algebra: an introduction |date=2014 |publisher=Brooks/Cole, Cengage Learning |isbn=978-1-111-56962-4 |edition=3rd |location=Boston, MA}}</ref>

The kernel is usually denoted {{nowrap|ker ''f''}} (or a variation). In symbols:
: <math> \ker f = \{g \in G : f(g) = e_{H}\} .</math>

Since a group homomorphism preserves identity elements, the identity element ''e''<sub>''G''</sub> of ''G'' must belong to the kernel.

The homomorphism ''f'' is injective if and only if its kernel is only the singleton set {''e''<sub>''G''</sub>}. If ''f'' were not injective, then the non-injective elements can form a distinct element of its kernel: there would exist {{nowrap|''a'', ''b'' &isin; ''G''}} such that {{nowrap|''a'' ≠ ''b''}} and {{nowrap|1=''f''(''a'') = ''f''(''b'')}}. Thus {{nowrap|1=''f''(''a'')''f''(''b'')<sup>−1</sup> = ''e''<sub>''H''</sub>}}. ''f'' is a group homomorphism, so inverses and group operations are preserved, giving {{nowrap|1=''f''(''ab''<sup>−1</sup>) = ''e''<sub>''H''</sub>}}; in other words, {{nowrap|''ab''<sup>−1</sup> &isin; ker ''f''}}, and ker ''f'' would not be the singleton. Conversely, distinct elements of the kernel violate injectivity directly: if there would exist an element {{nowrap|''g'' ≠ ''e''<sub>''G''</sub> &isin; ker ''f''}}, then {{nowrap|1=''f''(''g'') = ''f''(''e''<sub>''G''</sub>) = ''e''<sub>''H''</sub>}}, thus ''f'' would not be injective.

{{nowrap|ker ''f''}} is a [[subgroup]] of ''G'' and further it is a [[normal subgroup]]. Thus, there is a corresponding [[quotient group]] {{nowrap|''G'' / (ker ''f'')}}. This is isomorphic to ''f''(''G''), the image of ''G'' under ''f'' (which is a subgroup of ''H'' also), by the [[isomorphism theorems|first isomorphism theorem]] for groups.