A cyclic group is a group that is generated by a single element. In other words, every element in a cyclic group can be expressed as a power of a fixed element, called the generator.

To prove that every isomorphic image of a cyclic group is cyclic, we need to show that if two groups are isomorphic, then one being cyclic implies the other is also cyclic.


Let G be a cyclic group generated by an element a. This means that every element in G can be written as a power of a.

Let H be an isomorphic image of G, i.e., there exists an isomorphism f: G → H.

We want to show that H is also cyclic, i.e., there exists an element b in H such that every element in H can be written as a power of b.

Since f is an isomorphism, it is a bijective homomorphism. This means that for every element g in G, there exists a unique element h in H such that f(g) = h.

Let b = f(a). We claim that b generates H.

To prove this, let h be any element in H. Since f is surjective, there exists an element g in G such that f(g) = h.

Since G is cyclic, g can be written as a power of a, i.e., g = a^n for some integer n.

Now, using the properties of a homomorphism, we have:
f(g) = f(a^n) = f(a)^n = b^n

Therefore, h = b^n, which shows that every element in H can be written as a power of b.

Therefore, H is cyclic.


A homomorphic image of a group is a group that is obtained by applying a homomorphism to the original group.

To prove that every homomorphic image of a cyclic group is cyclic, we need to show that if H is a homomorphic image of G, where G is a cyclic group, then H is also cyclic.

However, this statement is false.


Consider the group G = Z (the integers under addition) which is cyclic and generated by the element 1.

Let H = {0, 2} which is a subset of Z.

Define a homomorphism f: Z → H by f(n) = 2n.

It can be shown that f is a well-defined homomorphism. However, the group H = {0, 2} is not cyclic. There is no element b in H such that every element in H can be written as a power of b.

Therefore, Statement B is false.

Therefore, the correct answer is option A: Both Statement A and Statement B are true.