Suppose $\phi: G \to G'$ is a homomorphism and $|G| = p$, $p$ is prime. Show either $\phi$ is injective or $\phi(g) = e_{G'}$ for all $g \in G$.

Question

Suppose $\phi: G \to G'$ is a homomorphism and $|G| = p$, $p$ is prime. Show either $\phi$ is injective or $\phi(g) = e_{G'}$ for all $g \in G$.

Attempt:

If $\phi(g) = e_{G'}$, we are done.

So assume $\phi(g) \ne e_{G'}$ and show that $\phi$ must be injective. So assume $\phi$ is not injective. This means, $\exists x,y \in G$ such that $\phi(x) = \phi(y)$. From Lagrange theorem, we have that since $|G| = p$ is prime, then $|x|, |y|$ divides $p$. So either $|x|,|y| = 1$ or $p$. Since neither $x,y$ are the identity, this means $|x| = |y|$.

I am not sure how to finish the proof that $x = y$.

Answer 1

You have the right idea, considering divisors of $p$ and their relationship to group theoretic properties.

Here's a hint for a slightly more elegant way to proceed:

Recall $K(\phi) = \{ g \mid \phi(g) = e_{G'} \}$ (the kernel of $\phi$) is always a subgroup of $G$. Then what are the possible sizes $|K(\phi)|$? Can you use the fact that $K(\phi) = \{ e_G \}$ if and only if $\phi$ is injective to finish the proof?

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I hope this helps ^_^