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Suppose that $G$ and $G'$ are two groups isomorphic to each other. Is it true that any onto homomorphism from $G$ to $G'$ is an isomorphism, i.e. any surjective homomorphism has a trivial kernel?

If it was not the case, then $G≈G'$ also Image of homomorphism $ =G'≈G/K$. Hence $G≈G/K$. This implies K is trivial. Am I right ?

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    It is true for finite groups, for trivial reasons. – 2017-01-10
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    Is it also true for groups with finitely countable elements ? @ Bernard – 2017-01-10
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    @ShreedharBhat no, my example works if you change $\mathbb R$ with a finite group. – 2017-01-10

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clearly false, consider $\mathbb R ^\mathbb N$ and the "scoot to the left" function. (in other words $(x_1,x_2,\dots)\rightarrow (x_2,x_3,\dots)$)

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The additive groups $(\mathbb C,+)$ and $(\mathbb R,+)$ are isomorphic; the map $z\mapsto\Re z$ is a surjective homomorphism but not an isomorphism.

Let $F$ be the free group generated by a countably infinite set $X.$ Any non-injective surjection from $X$ to $X$ extends to a surjective homomorphism from $F$ to $F$ which is not an isomorphism.