I am reading the proof of the 1-1 reasoning for the first homomorphism theorem. So, $G, G'$ are groups and there is a homomorphism $\phi: G \to G'$. $K$ is the kernel of $\phi$. The statement of the theorem is that $\psi: G/K \to G'$ defined by $\psi(Ka) = \phi(a)$ gives an isomorphism.

I understand that $\psi$ is a homomorphism and why it is onto. Intuitively, I see how it is 1-1 but I am not sure about the fine details. Here is what I have:

one-to-one: \begin{align*} \psi(Ka) =& \psi(Kb) \\ \phi(a) =& \phi(b) \\ \phi(a)\phi(b)^{-1} =& e\\ \phi(a)\phi(b^{-1}) =& e\\ \phi(ab^{-1}) =& e \\ ab^{-1} \in& K \\ ab^{-1} =& k_1k_2 \quad \text{for some $k_1,k_2\in K$}\\ k_3a =& k_2b \quad \text{where $k_3 = k_1^{-1}$}\\ \iff Ka =& Kb \end{align*}

I am unclear is my reasoning beyond $ab^{-1}\in K$

  1. Is this sufficient?
  2. What would you write?

Thanks for the help :)

  • $\begingroup$ Your steps after $ab^{-1} \in K$ are redundant and confusing. Just write then that $Kab^{-1} = K$ and so $Ka = Kb$. You can't make the 'if and only if' implication later down. Other than that I would write a similar proof. $\endgroup$ Dec 14, 2013 at 4:14

1 Answer 1


It is certainly sufficient, but over-complicated. $ab^{-1}\in K$ means $Kab^{-1}=K$. Your conclusion follows right from that!

  • $\begingroup$ Thanks for expounding on what I wrote in a comment; didn't think it warranted an answer though. $\endgroup$ Dec 14, 2013 at 4:19
  • $\begingroup$ Sorry, didn't see your comment. You're right about not warranting an answer, but I don't have the reputation points to comment yet! $\endgroup$
    – Nick D.
    Dec 14, 2013 at 4:24
  • $\begingroup$ Fair point, I didn't notice your reputation points (or lack thereof). +1 on good faith! $\endgroup$ Dec 14, 2013 at 4:52
  • $\begingroup$ ahhh, I see. $K$ multiplied to an element in $K$ is just $K$... Thanks! $\endgroup$ Dec 14, 2013 at 5:41

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