Let G be the direct sum of a countably infinite number of copies of Z. Find an element of End_Z(G) which has a left inverse, but is not a unit.

Please explain in detail.

Think of elements of End_Z(G) as infinite matrices with integer
entries.

Definition: Let G be an abelian group and let End_Z(G) be the set of all group homomorphisms from G to itself. We call it the endomorphism ring of G. Here, the ring operations are defined as follows. For f, g ∈ End_Z(G), we let f + g be the homomorphism given by (f + g)(x) = f(x) + g(x), where additive notation is used for the group operation in G. The product, f · g, of f, g ∈ EndZ(G) is simply the composition f ◦ g.

Lemma: Under these operations, End_Z(G) is a ring, with additive identity the trivial homomorphism, and multiplicative identity the identity map.

Find an element of End_Z(G) which has a left inverse, but is not a unit.

Solution Preview

The solution is attached.

We can think of G as a vector which has infinitely many components, each of which is an integer. Imagine taking a (row) vector and shifting all the elements1 position to the right to form a new vector, and filling in a 0 in the first position. ...

Solution Summary

The solution provides an example of finding an element with a left inverse.

... By:- Thokchom Sarojkumar Sinha. If the group has four elements, show it must be abelian. Solution:- Let , where is the identity element of . Order of . ...

... homomorphism f : G -> H is the set of all elements of G ... e) = e' (always), so that e is always an element of ker ... 1. If G is an Abelian group, then, for all a, b ...

Ring Theory : commutative ring with a unit element form abelian. Prove that the units in a commutative ring with a unit element form an abelian group. ...

... G ) and some integer m . Now we consider two elements x, y ... is generated by g and Z (G ) . Each element in H ... the proof in part (a), H is an abelian group, but Z ...

... coset (g^m)Z. Then we can rewrite x,y as x=(g^n)*z and y=(g^m)*z' for some z,z' in Z. Since elements in Z can commute with any element in G ... So G is abelian. ...

... table for the group and list the inverse elements i. For S ... show that the group is abelian 6. there are only ... tables for each one of them( the element should be ...

... Notes:- (a) Abelian Group A group is said to be abelian if for all . (b) Normalizer or Centralizer of an element of a Group Normalizer or Centralizer of an ...

... is the conjugacy class containing exactly 4 elements. ... if it's congugacy class contains exactly one element. ... only two non-isomorphic non-abelian groups of order ...

... Thus ab −1 moves at most m + n elements. ... TRUE. The cyclic group < g > generated by the element g in an infinite group G is a subgroup of G ... So G is abelian. ...

... Since this group is abelian, is a normal subgroup, and we can form the quotient group Z/ An element EZ/ is the equivalence class of numbers: [I4:] :I4:+(n ...