I've got a question on abstract algebra?Pls help?10 PTS!
[From: ] [author: ] [Date: 11-09-15] [Hit: ]
look that up online - theres tons of proofs.#3:A group must have: Closure, Associativity, Identity, and Inverseness. Closure: if a,......
INVERSE: If a is in subgroups H and K, then a must have an inverse in H and it must have an inverse in K. Since inverses are unique, a^-1 is in H and K, which means that every element in the intersection has an inverse.
Thus, we've shown that the intersection must be a valid subgroup. QED
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I don't really understand what #1 is sorry,
#2 is just a specific case of Fermat's Little Theorem, look that up online - theres tons of proofs.
#3:
A group must have: Closure, Associativity, Identity, and Inverseness.
Closure: if a,b are in G, then a+b and a*b are in G.
Proof: Let (a and b) £ (H inters. K).
Then obviously, a,b £H and a,b£K. According to the definition of subgroup, then a,b£G.
Then because every group must have closure, and G,H, and K are groups: (a+b) and (a*b)£G. (a+b) and (a*b)£H. (a+b) and (a*b)£K. By definition of intersection (a+b),(a*b)£(H intersection K)
Therefore we have proved that H intersection K has closure.
Indentity: for all a in G, there is a element i such that a*i=a and i*a=a.
Proof: Let b be an element in H intersection K. Therefore, b is an element of H and K, and G.
Since G is a group, there must be an identity element for b. So i£G. Since H and K also have b, and are also groups i£H and i£K. Therefore, i£(H intersection K)
Do you see the pattern? You just work your way up, showing that element b is in K intersection K, therefore it is in G. Because G is a group it has element c, and that element exists in H and in K, because they are subGROUPS. Since it exists in H and in K, it exists in H intersection K. Therefore, (after all four properties), H intersection K is a group, and obviously a subset of G. Igitur, H intersection K is a subgroup of G.
Q.E.D
Sorry I couldn't help on #1 :(
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