OpenStudy (anonymous):

Consider the group of quaternions Q = {1,-1,i,j,k,-i,-j,-k} a) Prove that Z(Q) = <-1> b) Build Q/Z(Q), listing its elements as cosets xZ(Q)= xbar, for example jZ(Q) = jbar. c) Prove that Q/Z(Q) ≅ K (Klein 4-group), by constructing the multiplication table of Q/Z(Q) and an isomorphism between K and Q/Z(Q).

6 years ago
OpenStudy (anonymous):

I think you need to attempt this yourself first. Also, the definitions of the quaternions and Z(Q) would be beneficial.

6 years ago
OpenStudy (anonymous):

Z(Q) is the center of the group, and I know it is {1,-1} but dont know how to prove that. For b), would the elements just be -1(1,-1,i,j,k,-i,-j,-k) = (1,-1,i,j,k,-i,-j,-k)bar? And i think i have the multiplication table for c), its a table of subgroups {1,-1},{i,-i},{j,-j}, and {k,-k}, is that right?

6 years ago
OpenStudy (anonymous):

The center of a group Z(G) is the set of elements that commute with every element in G. Show you show that for each g in G, 1*g = g*1. That is true by definition of 1. Then for each g in g, -1*g = g*(-1). That should also be simple enough. Then for each other element, you can show an example where z*g =! g*z. (=! means does not equal).

6 years ago
OpenStudy (anonymous):

Q/Z(Q) is the set of all cosets xZ(Q) = x{-1,1}. Now for each x in G, compute what x{-1,1} would be (you know, {-1*x, x}). Some of them will be repeats. Don't list them twice in your final answer.

6 years ago
OpenStudy (anonymous):

So the subgroups you wrote for c) is indeed the answer to b), but of course you must show your work. As for the isomorphism in c), remember that is a 1-1, onto, and a homomorphism. A homomorphism satisfies f(x*y) = f(x)$f(y), where * is the multiplication of two elements x and y in K, and $ is the multiplication of two elements in Q/Z(Q).

6 years ago
OpenStudy (anonymous):

Awesome, you explained that very clearly thank you.

6 years ago