Let G and G' be finite groups and $\phi$: G $\rightarrow$ G' a homomorphism.
(a) Suppose that |G'| is prime. Show that $\phi$ must either be trivial or surjective.
(b) Suppose that |G| is prime. Show that $\phi$ must either be trivial or injective.
(c) Suppose that |G| and |G'| are the same prime number. Show that $\phi$ is either trivial or an isomorphism.
(d) Suppose that |G| and |G'| are relatively prime. Show that $\phi$ must be trivial.
My approach for (a) is: Since |G'| is prime. The the only two subgroups of G' must be the id. element and G' itself. Let H be a subgroup of G, then $\phi (H)$ must be a subgroup in G'. Then $\phi (H)$ must be either the id. element or the G'. Therefore, it's trivial or surjective. Is this correct??
for (b), could I do it with the same approach: |G| is prime. Then the subgroups are id. element and G itself. Define H to be the subgroup of G. But I do not how to go from here.
for (c). Both |G| and |G'| are prime. Then the subgroups of G are id. element and G. For G', its {e'} and G'. If H (H is subgroup of G) is {e}, then $\phi$ is trivial. If H is G, then it maps G to G' which is one to one and onto. I wonder how can I prove its one-to-one and onto here.
for (d). I really don't know what to start with. Should go with the same approach?
a. You're on the right track. $\phi(G)$ is a subgroup of $G^{\prime}$, hence is either trivial or all of $G^{\prime}$.
b. Try the same argument as (a), but now applied to the subgroup $\mathrm{ker}(G)$ of $G$.
c. Combine (a) and (b).
d. $\phi(G)$ is a subgroup of $G^{\prime}$, so its order divides $|G^{\prime}|$. But it's also a quotient of $G$, so its order divides $|G|$. Since $|G|$ and $|G^{\prime}|$ are coprime, this means that $\phi(G)$ is trivial.