let a AB and BBC be homomorphisms the we have a Bare efimorphisms Bod is ae Pimorphisms pod is epimorphism Befimorphism let SIA B be amodule homemphism and let u besub Module of A It is well-know that UCS (S(u)) When U = $($(u) and why? Compare be tween Maximal submodule and largest submodules. The trive submodule is the only Smal submodule in Zz prove.

need help with this question modern algebra dealing with cosets

Subject Test Which of the following functions are the isomorphisms? a) :(Z,+)→(Z,+) defined by o(n) = 2n, for xeZ %3D b) :(Q+)→(Q.+) defined by ø(x) =, for IEQ c) :(Z,+)→(Z,+) defined by (n) =n+1, for xeZ d) :(Q.-)→(Q.-) defined by o(x)= x, for xeQ Answer B C.

Abstarct Algebra help please..need to understand

5. Prove analogues of the results in problems 2 and 3 for arbitrary unions and intersections U₁4 and Ax

Parts (b,c,d,e) please

whats the answer ? thanks

Describe the kernel of epimorphism in Exercise 20. Consider the mapping :Z[ x ]Zk[ x ] defined by (a0+a1x++anxn)=[ a0 ]+[ a1 ]x++[ an ]xn, where [ ai ] denotes the congruence class of Zk that contains ai. Prove that is an epimorphism from Z[ x ] to Zk[ x ].

True or False Label each of the following statements as either true or false. Disjoint cycles commute under mapping composition.

Let G=1,i,1,i under multiplication, and let G=4=[ 0 ],[ 1 ],[ 2 ],[ 3 ] under addition. Find an isomorphism from G to G that is different from the one given in Example 5 of this section. Example 5 Consider G=1,i,1,i under multiplication and G=4=[ 0 ],[ 1 ],[ 2 ],[ 3 ] under addition. In order to define a mapping :G4 that is an isomorphism, one requirement is that must map the identity element 1 of G to the identity element [ 0 ] of 4 (part a of Theorem 3.30). Thus (1)=[ 0 ]. Another requirement is that inverses must map onto inverses (part b of Theorem 3.30). That is, if we take (i)=[ 1 ] then (i1)=((i))1=[ 1 ] Or (i)=[ 3 ] The remaining elements 1 in G and [ 2 ] in 4 are their own inverses, so we take (1)=[ 2 ]. Thus the mapping :G4 defined by (1)=[ 0 ], (i)=[ 1 ], (1)=[ 2 ], (i)=[ 3 ]