You are not logged in.
Pages: 1
I don't understand why the intersection of groups is still a group ?
Numbers are the essence of the Universe
Offline
here is the question Prove that the intersection of a family of subgroups of a group G is a subgroup of G.
Numbers are the essence of the Universe
Offline
and another one , Let a,b,c,d, be integers such that ad-bc=1 , for integer u, v define
m=au+bv , n=cu+dv . Prove that (m,n) = (u,v)
Last edited by Stanley_Marsh (2007-01-24 13:20:38)
Numbers are the essence of the Universe
Offline
Yes, this can be a bit confusing at first. Doesn't seem like it should be true, right? But, just run through all the properties:
Let A and B be two subgroups of G. We let C signify A intersect B.
e must be in A, and e must be in B, so e must be in C. Thus, C contains the identity.
Now we let some two elements x and y be in C. Then x is in A and x is in B and y is in A and y is in B. So xy is in A as A must be closed, and for the same reason, xy is in B. So xy is in C, and C is closed.
Now let x be in C. So x is in A and x is in B. Since A and B are closed to inverses, x inverse is in A and x inverse is in B. So x inverse is in C, and thus, C is closed ot invserses.
Finally, we can just say that C is a subset of G, and thus, all elements in C must be associative, as all elements in G are.
And that's it. Also note that we can expand C to be the intersection of A_1, A_2, A_3, ..., A_n, and all the above statements still apply.
"In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..."
Offline
Oh ,yeah, I can assume x, y in C , instead I assume them in other wrong places , It's really confusing at the beginning ~
Numbers are the essence of the Universe
Offline
Pages: 1