Prove that a homomorphism is injective or trivial

Question

Let A,B be groups, and assume that |A| = 29. Let φ:A→B be a homomorphism.

a) Prove that either φ is injective or trivial. (φ is trivial if for all a∈A φ(a) = e)

b) If |B|=80, prove that φ is trivial.

Now I know that a homomorphism is injective iff the kernel is trivial. But I can't seem to figure out how to start this question. Should I assume by contradiction that φ is not injective and not trivial and try to arrive at a contradiction? Or should I show that the kernel is trivial so φ has to be injective.

Any hints or suggestions would be really appreciated. Thanks!

Answer 1

Hints:

• The kernel of $\phi$ is a subgroup of $A$.
• The order of a subgroup must divide the order of the group.
• The image of $\phi$ is a subgroup of $B$.

Answer 2

By the first isomorphism theorem, $A/\ker\varphi \cong \varphi(A)$. In particular: \begin{equation}\frac{29}{|\ker\varphi|}=|\varphi(A)|.\end{equation} Since 29 is a prime number, either $\ker\varphi=\{0\}$ (which implies $\varphi$ is injective) or $\ker\varphi=A$ (i.e. $\varphi$ is trivial).

If $|B|=80$ then since $\varphi(A)$ is a subgroup of $B$, $|\varphi(A)|$ divides $80$ by Lagrange's theorem. But $|\varphi(A)|$ also divides 29, so $|\varphi(A)|=1$ and $|\ker\varphi|=29$, which implies $\varphi$ is trivial.