Exact sequence of groups: proof of injectivity

Question

There must be a duplicate being the question very introductory, but I was not able to find it.

We have the following diagram

$$\begin{array}{ccccccccc} 1&\to& H &\to & G &\overset{\pi}{\to} & G/H &\to & 1\\ && \downarrow & & \downarrow f && \downarrow g & & \\ 1&\to& H' &\to & G'& \overset{\pi'}{\to} & G'/H' &\to & 1\\ \end{array}$$

Suppose that

1. $f$ is homomorphism
2. $H'\triangleleft G'$
3. $H=f^{-1}(H')$
4. $H\triangleleft G$
5. $f$ is surjective

Prove that $g$ is an isomorphism.

Since $g\circ\pi=\pi'\circ f$ and $\pi,\pi',f$ are surjective, then $g$ is surjective.

The problem is with injectivity. (A) From $g(aH)=g(bH)$ I should derive $aH=bH$. Or (B) prove that $\operatorname{ker} g=\{H\}$.

(A) Working backwards I get $$ aH=bH \iff a^{-1}b\in H \iff f(a^{-1}b)\in H'\\ \iff (f(a))^{-1}f(b)\in H' \iff f(a)H'=f(b)H' $$

Now, if $g(aH)=g(bH)$ how do I get $f(a)H'=f(b)H'$?

Method (B) would probably be nicer, but no idea how to start.

Answer 1

To prove directly that $\DeclareMathOperator\Ker{Ker}\Ker g$ is trivial, note that \begin{align} \pi^{-1}\Ker g &=\Ker(g\circ\pi)\\ &=\Ker(\pi'\circ f)\\ &=f^{-1}\Ker\pi'\\ &=f^{-1}H'\\ &=H \end{align} Since $\pi$ is surjective, we get $\Ker g=\pi[\pi^{-1}\Ker g]=\pi[H]=\{1\}$.