About Group Isomorphism Kernel It's clear to me that a group homomorphism kernel measures how much the homomorphism is close to be an isomorphism, i.e. if this kernel contains more than the identity, the homomorphism cannot be injective, and so cannot lead to any isomorphism at all.
This sounds to me a "necessary" condition, so far so good.
I was wondering whether this is a "sufficient" condition as well, i.e. whether to check a kernel being made by the identity only it's ENOUGH to claim the homomorphism is actually a isomorphism.
Please, give some counter examples in negative case.
Tx in advance.
 A: For finite groups since $f$ is injective if and only if $Kerf=\{0\}$. If We have $f : G\to H$ with $ord(G) =ord(H) $, f is isomorphism if and only if the kernel is trivial. For infinite groups that's not true, for example $f : (\mathbb{Z},+) \to (\mathbb{Z}, +), \quad x\mapsto 2x$
A: Let $\varphi : G\to G'$ be a group homomorphism.
Then $\ker(\varphi) $ measure how far a homorphism to be an injective map.
Injectivity alone isn't enough to ensure whether the homomorphism is an isomorphism Or not.
Example : $ \varphi: S_n \to GL_n(\Bbb{Z}) $ defined by $\varphi(\pi) =P_{\pi}$ where $P_{\pi}=(e_{\pi(1)},e_{\pi(2)}, \ldots e_{\pi(n)})$ and $e_j$ is the $j-$th standard basis vector (column) .
Then $\ker(\varphi)=\{(1)\}$ but $\varphi$ isn't an isomorphism.Find $\varphi(S_n) $ ?
A injective homomorphism $\varphi : G\to G'$  is an isomorphism iff $\varphi(G) $ is an improper subgroup of $G'$  iff $\varphi$ is onto.
A: The kernel being trivial is equivalent to the homomorphism being injective, but it may still fail to be bijective if it is not surjective. Take $C_2\to C_4\colon a\mapsto b^2$ where $a$ and $b$ are generators of the cyclic groups $C_2$ and $C_4$ respectively.
What you can say is that, given a homomorphism $f\colon G\to H$, then the kernel of $f$ is trivial if and only if $f$ defines an isomorphism between $G$ and its image $f(G)$, a subgroup of $H$.
