Find homomorphisms from z8 to Q8 ?
Answers
Answer:
We have (Q_8) = {1, -1, i, -i, j, -j, k, -k} and (Z_4)={[0],[1],[2],[3]}.
If f: (Q_8)→(Z_4) is a homomorphism, then f(1) = [0], f(-1)=[0] or [2] because [(-1)^2]=1.
In the first case, the order of f(i) must divide the order of i, which is 4 and so it f(i) is of order 4, 2 or 1. In the first case f(i) can be [1] or [3]. Accordingly f(-i) = [3] or [1] respectively. And in either case f(-1) = f(i.i)= [2].
Next if we also take f(j) = [1], then f(k)=f(i.j) = f(i)+f(j) = [1]+[1]=[2] or [3]+[1]=[0]. This gives [2] = f(-1) = f(k.k) = f(k)+f(k) =[2]+[2]=[0] or [0]+[0] =[0], leading to a contradiction. This implies that both f(i) and f(j) cannot be of order 4.
Therefore we take f(j)=[2]. This gives [2] = f(-1)=f(j.j)=f(j)+f(j)=[2]+[2]=[0] #. The above arguments show that none of the elements of Q_8, having order 4, can be mapped into an element of order 4 in Z_4.
Thus the images of i, j and k and their inverses must be [2] or [0]. If f(i)=[2]=f(j), then f(k)=f(i.j)=f(i)+f(j)=[2]+[2]=[0] =f(-k). A bit of reflection in consideration of the structure of Q_8 about symmetry of the elements of order of 4 shows that the only homomorphims: Q_8→Z_4 are as follows:
1: f(a) = [0] for all a in Q_8.
2: f = {(1, [0]), (-1, [0]), (i, [2]), (-i, [2]), (j, [2]), (-j, [2]), (k, [0]), (-k, [0])}.
3: f = {(1, [0]), (-1, [0]), (i, [2]), (-i, [2]), (j, [0]), (-j, [0]), (k, [2]), (-k, [2])}.
4: f = {(1, [0]), (-1, [0]), (i, [0]), (-i, [0]), (j, [2]), (-j, [2]), (k, [2]), (-k, [2])}.
These are all the distinct homomorphisms from Q_8 to Z_4.
We now find the homomorphism from the Hamiltonian group Q_8 to the group Z_8 of integers under addition modulo 8:
Z_8 = {[0], [1], [2], [3], [4], [5], [6], [7]}. Note that this group has the elements [1], [3], [5] and [7] which are of order 8 as these are each a generator of the group. Since the orders of the elements of Q_8 are 1, 2 or 4 only, none of their elements can be mapped by a homomorphism. Hence if f:Q_8→Z_8 is any homomorphism, the image of f is contained in the subgroup H={[0], [2], [4], [6]}. Also note that H is isomorphic to the group Z_4 under +. Hence we can obtain all the homomorphisms by the first half of the answer simply by replacing the image [2] there by [4] here, as H is a cyclic group of order 4 generated by [2] (or by [6]). Hence there are exactly 4 homomorphisms here also which can be written down explicitly, as indicated above.