proof of properties of Hopfian and co-Hopfian groups

Proposition. A group $G$ is Hopfian if and only if every surjective homomorphism $G\to G$ is an automorphism.

Proof. “$\Rightarrow$” Assume that $\psi :G\to G$ is a surjective homomorpism such that $\psi$ is not an automorphism, which means that $\mathrm{Ker}(\psi )$ is nontrivial. Then (due to the First Isomorphism Theorem) $G/\mathrm{Ker}(\psi )$ is isomorphic to $\mathrm{Im}(\psi )=G$. Contradiction, since $G$ is Hopfian.

“$\Leftarrow$” Assume that $G$ is not Hopfian. Then there exists nontrivial normal subgroup $H$ of $G$ and an isomorphism $\varphi :G/H\to G$. Let $\pi :G\to G/H$ be the quotient homomorphism. Then obviously $\pi \circ \varphi :G\to G$ is a surjective homomorphism, but $\mathrm{Ker}(\pi \circ \varphi )=H$ is nontrivial, therefore $\pi \circ \varphi$ is not an automorphism. Contradiction. $\mathrm{\square }$

Proposition. A group $G$ is co-Hopfian if and only if every injective homomorphism $G\to G$ is an automorphism.

Proof. “$\Rightarrow$” Assume that $\psi :G\to G$ is an injective homomorphism which is not an automorphism. Therefore $\mathrm{Im}(\psi )$ is a proper subgroup of $G$, therefore (since $\mathrm{Ker}(\psi )=\{e\}$ and due to the First Isomorphism Theorem) $G$ is isomorphic to its proper subgroup, namely $\mathrm{Im}(\psi )$. Contradiction, since $G$ is co-Hopfian.

“$\Leftarrow$” Assume that $G$ is not co-Hopfian. Then there exists a proper subgroup $H$ of $G$ and an isomorphism $\varphi :G\to H$. Let $i:H\to G$ be an inclusion homomorphism. Then $i\circ \varphi :G\to G$ is an injective homomorphism which is not onto (because $i$ is not). Contradiction. $\mathrm{\square }$