Ask any question about EPFL courses, lectures, exercises, research, news, etc. or try the example questions below.
DISCLAIMER: The Graph Chatbot is not programmed to provide explicit or categorical answers to your questions. Rather, it transforms your questions into API requests that are distributed across the various IT services officially administered by EPFL. Its purpose is solely to collect and recommend relevant references to content that you can explore to help you answer your questions.
In abstract algebra, the fundamental theorem on homomorphisms, also known as the fundamental homomorphism theorem, or the first isomorphism theorem, relates the structure of two objects between which a homomorphism is given, and of the kernel and of the homomorphism. The homomorphism theorem is used to prove the isomorphism theorems. Given two groups G and H and a group homomorphism f : G → H, let N be a normal subgroup in G and φ the natural surjective homomorphism G → G/N (where G/N is the quotient group of G by N). If N is a subset of ker(f) then there exists a unique homomorphism h: G/N → H such that f = h∘φ. In other words, the natural projection φ is universal among homomorphisms on G that map N to the identity element. The situation is described by the following commutative diagram: h is injective if and only if N = ker(f). Therefore, by setting N = ker(f) we immediately get the first isomorphism theorem. We can write the statement of the fundamental theorem on homomorphisms of groups as "every homomorphic image of a group is isomorphic to a quotient group". The proof follows from two basic facts about homomorphisms, namely their preservation of the group operation, and their mapping of the identity element to the identity element. We need to show that if is a homomorphism of groups, then: is a subgroup of . is isomorphic to . The operation that is preserved by is the group operation. If , then there exist elements such that and . For these and , we have (since preserves the group operation), and thus, the closure property is satisfied in . The identity element is also in because maps the identity element of to it. Since every element in has an inverse such that (because preserves the inverse property as well), we have an inverse for each element in , therefore, is a subgroup of . Construct a map by . This map is well-defined, as if , then and so which gives . This map is an isomorphism. is surjective onto by definition. To show injectiveness, if , then , which implies so . Finally, hence preserves the group operation.
[2]Paolo Prandoni, [3]LCAV, EPFL
The concept of "basis" for a vector space is one of the most profound
and fruitful ideas in linear algebra. Basis vectors are the "building
blocks" that all other vectors are built from; by "disassembling" a
vector into a linear combination of basis elements we are able to "look
inside" the vector and highlight many of its relevant features.
A vector space has an infinite number o