Mapping between two cyclic groups

• Feb 16th 2011, 07:15 AM
Zalren
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?
• Feb 16th 2011, 07:20 AM
roninpro
Try using the fact that a homomorphism takes subgroups to subgroups. What are the subgroups of the cyclic group of order 7?
• Feb 16th 2011, 07:50 AM
Zalren
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?
• Feb 16th 2011, 07:53 AM
roninpro
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.