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I have this proposition:

A group homomorphism $\varphi: G \sim G'$ is injective if and only if its kernel is the trivial subgroup {1}.

This is my proof:

We assume a homomorphism is injective: Given $a,b\in G$, $\varphi(a)=\varphi(b)$ only when $a= b$.

Let $a,b\in G$, $N = ker(\varphi)$. So to make $\varphi(a)=\varphi(b)$, we need $\varphi(aN)=\varphi(bN)$.

As per the assumption, $aN=bN$. And if kernel $N$ is not trivial, we can find different $n_{1},n_{2}\in N$, which lead to $an_{1}\neq an_{2}$.

Is this correct? If not could you give me a proof? And can I rigorously say "if $n_{1}\neq n_{2}$, then $an_{1}\neq an_{2}$"? Is there any guarantee for this in group theory? Thanks.

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1 Answer 1

up vote 1 down vote accepted

You are overcomplicating matters. Kernel trivial is equivalent to saying that the identity has exactly one preimage. You want to show that EVERY element has only one preimage (that's what it means to be injective). So, if $\phi(a) = \phi(b),$ then $\phi(ab^{-1}) = 1,$ so $a b^{-1}$ is in the kernel.

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Aha, so then a=b! I really did complicate it a lot... Thank you! –  babel92 Jan 6 at 20:52

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