# Thread: Mapping between two cyclic groups

1. ## Mapping between two cyclic groups

Let $\displaystyle G$ be a cyclic group of order 5. Let $\displaystyle G'$ be a cyclic group of order 7. Let $\displaystyle f : G -> G'$ be a homomorphism. Why is $\displaystyle f$ the trivial homomorphism?

In my group, the trivial homomorphism means the map that takes every element of the domain and sends it to the $\displaystyle e$ element of the codomain.

I know what $\displaystyle G$ and $\displaystyle G'$ look like.

$\displaystyle G = \{e, a, a^2, a^3, a^4\}$ and $\displaystyle G' = \{e, b, b^2, b^3, b^4, b^5, b^6\}$

But how is it that $\displaystyle f$ must be the trivial homomorphism?

2. Try using the fact that a homomorphism takes subgroups to subgroups. What are the subgroups of the cyclic group of order 7?

3. I found this theorem in my book. Let $\displaystyle f : G -> G'$ be a group-homomorphism. The image of $\displaystyle f$ is a subgroup of $\displaystyle G'$.

So by this theorem, $\displaystyle f(G)$ is a subgroup of $\displaystyle G'$.

Then I found this online: Let a cyclic group have order n. If r is a divisor of n then there is some number s such that rs = n. Let a be an element of the group with order n. (it must have one if it is cyclic) Then ar is of order s and therefore generates a cyclic subgroup of order s while as generates a subgroup of order r.

So if that is true, since 7 is a prime the only subgroups $\displaystyle G'$ has is of order 1 or 7. This would imply that $\displaystyle f$ is the trivial homomorphism.

Is this correct?

4. That's right. You have to send $\displaystyle G$ to $\displaystyle G'$ or $\displaystyle \{e'\}$. The former isn't possible because it is too large, so $\displaystyle G\mapsto \{e'\}$, which is the trivial homomorphism.

Nicely done.