Prove that if $\varphi$ is injective then $G \cong \varphi(G)$

Question

Let $G$ and $H$ be groups and let $\varphi : G \to H$ be a homomorphism and the image of $\varphi$ , $\varphi(G)$ is a subgroup of $H$ . Prove that if $\varphi$ is injective then $G \cong \varphi(G)$

My attempt:

If $\varphi$ is injective, then for every $a \in \varphi(G)$,there exist $b \in G$ such that $\varphi(b)=a$.

After that I'm not able to proceed any further.

Answer 1

Use the first isomorphism theorem and the fact that $\varphi$ is injective iff $\ker\varphi$ is trivial.

Answer 2

Hint:

$$G/\ker(\phi)\cong\phi(G)$$

Answer 3

Just in case you haven't got the first homomorphism theorem at hand. Any map $f\colon X\to Y$ gives rise to a surjective (by construction) map $\tilde f\colon X\to f(X)$ defined by $\tilde f(x):=f(x)$. Therefore, your $\varphi$ gives rise to a bijective homomorphism ($=$isomorphism) $\tilde \varphi\colon G\to \varphi(G)$, meaning $G\cong\varphi(G)$.

Answer 4

Define $f:G\to \varphi(G)$ such that $f:g\mapsto \varphi(g)$. Clearly, $f$ is surjective. We are given that $\varphi$ is injective, hence $f$ is injective. So, $f$ is a bijection. Since $\varphi$ is a homomorphism, so is $f$. Thus, $G\cong\varphi(G)$.