Let o: G1G2 be a surjection homomorphism between two groups, N be a normal subgroup of gi, and KerocN. Proof: G2/(N)is a group. Theorem 6.22: Let [H; be a normal subgroup of the group G; * Then G/H; o is a group. VP(N is a subgroup of g2. ◆ Normal subgroup? ◆g2op(n)g2∈?(N) ◆ For vg2∈G2,p(n)∈q(N),g1∈G1S.t q(g1)=g2 o surjection homomorphism
Let : G1→G2 be a surjection homomorphism between two groups, N be a normal subgroup of G1 , and KerN. Proof: G2 /(N) is a group. Theorem 6.22: Let [H;] be a normal subgroup of the group [G;]. Then [G/H;] is a group. (N) is a subgroup of G2 . Normal subgroup? g2 -1(n)g2?(N) For g2G2 , (n)(N), g1G1 s.t (g1 )=g2 ? surjection homomorphism
Let o: G1G2 be a surjection homomorphism between two groups, n be a normal subgroup ofG1, and Kerc∈N. Proof:G1N≌G2/@p(N) Corollary 6.2: If op is a homomorphism function from group G; to group G; and it is onto, then IG/K;lG’; ◆ Prove:f:G1→→G2/(N) ◆ For Vg1∈G12f(g1)=(N)p(g1 Ifis a surjection homomorphism from Gi to 2/@(N ◆2)kerf=N
Let : G1→G2 be a surjection homomorphism between two groups, N be a normal subgroup of G1 , and Ker N. Proof: G1 /N≌G2 /(N). Corollary 6.2: If is a homomorphism function from group [G;*] to group [G';•], and it is onto, then [G/K;][G';•] Prove:f :G1→G2 /(N) For g1G1 , f (g1 )=(N)(g1 ) 1)f is a surjection homomorphism from G1 to G2 / (N). 2)Kerf =N
6.6 Rings and fields 6.6.1 Rings 6 Definition 21: A ring is an Abelian group r,+ with an additional associative binary operation (denoted such that for all a, b, CER, (1)a·(b+c)=a·b+a·c, (2)(b+c)·a=b·a+c·a We write oER for the identity element of the group R,+ t Fora eR. we write -a for the additive inverse of a o Remark: Observe that the addition operation is always commutative while the multiplication need not be e Observe that there need not be inverses for multiplication
6.6 Rings and fields 6.6.1 Rings Definition 21: A ring is an Abelian group [R, +] with an additional associative binary operation (denoted ·) such that for all a, b, cR, (1) a · (b + c) = a · b + a · c, (2) (b + c) · a = b · a + c · a. We write 0R for the identity element of the group [R, +]. For a R, we write -a for the additive inverse of a. Remark: Observe that the addition operation is always commutative while the multiplication need not be. Observe that there need not be inverses for multiplication
Example: The sets Z,Q, with the usual operations of multiplication and addition form rings, [Z;+,×],Q+,× are rings Let m=((aiinxnlai is real number, Then M; +, x is a ring ◆ Example:S≠,P(S):,n], ◆ Commutative ring
Example: The sets Z,Q, with the usual operations of multiplication and addition form rings, [Z;+,],[Q;+,] are rings Let M={(aij)nn |aij is real number}, Then [M;+,]is a ring Example: S,[P(S);,∩]， Commutative ring
Definition 23: A ring r is a commutative ring if ab= ba for all a, bER. A ring r is an unitary ring if there is lER such that la= al a for all aeR. Such an element is called a multiplicative identity
◆Definition 23: A ring R is a commutative ring if ab = ba for all a, bR . A ring R is an unitary ring if there is 1R such that 1a = a1 = a for all aR. Such an element is called a multiplicative identity