If two (right) cosets share an element, then they are equal.

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• Index of subgroup and coset of subgroup

  Specifically, there are 2 right cosets of N in G (namely, N and N*a for some element "a" which is not an element of N), and there are two left cosets of N in G (namely, N and a*N).

• Normal subgroups and product of two right cosets

  By the definition, any two right cosets of in either are identical or have no element in common.

• Minimal polynomial and stabilizer

  For any two elements , where , we have . Then we have . So the two elements maps to the same element in . Second, for two different cosets and , I claim that . Because if , then . So we have . This is a contradiction.

• Groups and proofs

  We note, for any two right cosets and in , either or . This means any two right cosets are either equal or disjoint. Since is a subgroup, then , then we have and Since , then . Thus . So and are not disjoint right cosets.

• Abstract Algebra: Identity Element of the Group

  Define two elements of this set to be related if their derivatives are equal. Prove that this defines an equivalence relation. 3. Let H be a subgroup of the group G. Prove that every right coset of H is a left coset of some subgroup of G. 1.

• Cosets

  For any left coset bH, we can find a right coset Hb^(-1) such that $(Hb^(-1))=(b^(-1))^(-1)H=bH. Third, $ is one-to-one. If Ha and Hb are two different right cosets, then $(Ha)=a^(-1)H, $(Hb)=b^(-1)H.

• Ideals : Cyclic Module in a Commutative Ring

  But f is NOT a ring homomorphism since f(st) is not equal to f(s)*f(t) Let L = ker f = {r element of R such that ra = 0} L is an ideal in R: First, L is a subring; 0 is an element of L; if s,t are in L then (s + t)a = sa + ta = 0 + 0 = 0 so s+t

• Some concepts of groups and rings

  Let us construct the mapping from the set to the set of right cosets of by . Consider . There exists such that . We send to . This mapping is defined correctly. Indeed, if , then Therefore, . Hence .

• Infinite Groups

  Now if you take a rational number r/s, where r and s are integers, then s (r/s) = r which is an integer.