1. let H be a subgroup of a group G such that g ֿ ¹ hg elements in H for all h elements in H. Show every left coset gH is the same as the right coset Hg.
2. prove that if G is an abelian group, written multiplicatively, with identity element e, then all elements, x, of G satisfying the equation x²=e form a sub group H of G
3. show that if a elements in G where G is a finite group with the identity, e, then there exist n elements in Z+ such that a ⁿ =e
4. prove the generalisation of the first part of this question: consider the set H of all solutions, x, of the equation x ⁿ =e for fixed integer n ≥1 in an abelian group, G with identity , e.
5. if ◦ is a binary operation on a set, S an element, x elements in S is an idempotent for ◦ if x ◦ x= x. prove that a group has exactly one idempotent element.
6. define the term 'normal subgroup'. Give an example of a group, G, and a normal subgroup, H, of G
7. prove that every group, G, with identity, e, such that x ◦ x=e for all x G is abelian .
8. draw the cayley tables for the Z and V. for each group, list the pairs of inverses
9. determine whether the following are hohmorphisms. Let:
i. ϕ : Z → R under addition be given by ϕ (n) =n.
ii let G be any group and let : G→ G be given by ϕ (g)=g ֿ ¹
for g elements in G
10. show that if G is nonablelian, the quot6ient group G/Z(G) is not cyclic ( i know that some how i have to show the equivalent contrapositive, ie that if G/Z(G) is cyclic then g is ablelian and hence Z(G)=G)
11. show that the intersection of normal subgroups of a group G is again a normal subgroup of G
First, I claim that gH is contained in Hg. For each element gh in gH, since ghg^(-1)=r^(-1)hr, where r=g^(-1), then ghg^(-1) is an element of H. So we can find some h' in H, such that ghg^(-1)=h'. Then gh=h'g is an element in Hg. So gH is contained in Hg.
Second, I claim that Hg is contained in gH. For each element hg in Hg, since g^(-1)hg is an element in H, then we can find some h' in H, such that g^(-1)hg=h'. So hg=gh' is an element in gH. Thus Hg is contained in gH.
Suppose H is the subset of G such that for any x in H, x^2=e. Now I show that H is a subgroup of G. For any x,y in H, we know x^2=y^2=e. Then y^(-1)^2=y^2^(-1)=e^(-1)=e. Since G is abelian, then (xy^(-1))^2=x^2*y^(-1)^2=e*e=e. So xy^(-1) is an element in H. Therefore, H is a subgroup of G.
G is a finite group, we can suppose |G|=k. For any element a in G, we consider the following k+1 elements: a,a^2,a^3,...,a^k,a^(k+1). All of them belong to G. But G has only k ...
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