Solution

The correct answer is An onto homomorphism, which is not one to one

Key Points

• The map ϕ: G → H is defined by ϕ(x) = x / |x|. G here is the set of non-zero real numbers and H is the set {-1, 1}. We are considering these groups under multiplication.
• Now, a mapping φ: G → H is called a homomorphism if for all a, b in G, we have ϕ(ab) = ϕ(a) * ϕ(b).
• Our mapping here, ϕ(x) = x / |x|, takes any positive number in G to 1 in H and any negative number in G to -1 in H. So, if we take two elements a and b in G, we have:
• ϕ(ab) = ab / |ab| ϕ(a) * ϕ(b) = (a / |a|) * (b / |b|)
• If a and b are both positive or both negative, |ab| = |a||b| so ϕ(ab) = ϕ(a) * ϕ(b). If a and b are of opposite signs, |ab| = |a||b| holds as well, therefore ϕ(ab) = ϕ(a) * ϕ(b) is true in this case, too.

This shows that the mapping ϕ is a homomorphism.

check whether it is one-one (injective) or onto (surjective).

• The function is not one-one, because multiple distinct elements in G (all positive numbers, for example) get mapped to the same element in H (1 in this case).
• The function is onto, because every element in H can be reached by some element in G (positive numbers map to 1 and negative numbers map to -1).

So, applying these conclusions to the options, the answer is:

3) An onto homomorphism, which is not one to one.