# Proving that the kernel of a homomorphism is exactly equal to a normal subgroup $A'$

Here is the background for this problem.

Let $$\phi: A \to B$$ be a group homomorphism, $$A' \subset A$$ a normal subgroup where $$A' \subset \mathrm{ker}(\phi)$$, and $$\pi: A \to A/A'$$ the canonical quotient map. There exists a unique group homomorphism $$\tilde{\phi}: A/A' \to B$$ such that $$\phi = \tilde{\phi} \circ \pi$$.

I was already able to prove this fact. The group structure on $$A/A'$$ is "multiplication of cosets" and the map $$\tilde{\phi}$$ is defined by $$\tilde{\phi}(y) = x$$ where $$\pi(x) = y$$. I proved that this map is well-defined and satisfies all of the required properties. I am trying to prove the following corollary.

Show that $$\tilde{\phi}$$ is injective if and only if $$A' \subset \mathrm{ker}(\phi)$$ is an equality.

I'm fine with the backward direction. The forward direction I am struggling with. Here is my attempt.

Suppose $$x \in \mathrm{ker}(\phi)$$, so $$\phi(x) = 1_B$$. We need to show that $$x \in A'$$. (I believe this entails showing that $$x = 1_A$$ in which case $$x \in A'$$ because $$A'$$ is a subgroup.) Because $$\tilde{\phi}$$ is injective, its kernel is trivial, so $$\mathrm{ker}(\tilde{\phi}) = \{\overline{1_A}\}$$. Then $$1_B = \phi(x) = (\tilde{\phi} \circ \pi)(x) = \tilde{\phi}(\pi(x)).$$ As $$\tilde{\phi}$$ has a trivial kernel, $$\pi(x) = \overline{1_A} = \overline{x}$$, so $$x \sim 1_A$$ under the equivalence relation on the fibres of $$\phi$$, so $$\phi(x) = \phi(1_A) = 1_B$$.

This produced a statement I already knew to be true, however, as $$\phi$$ was a homomorphism and therefore must carry the identity to the identity. This doesn't establish that $$\phi$$ is injective which, while not true, would imply the result.

I would appreciate a hint on how to proceed.

If there was some $$x\in Ker(\phi)\setminus A'$$ then we would have $$\tilde{\phi}(xA')=\phi(x)=e_B$$, while $$xA'\ne A'$$. This contradicts the assumption that the kernel of $$\tilde{\phi}$$ is trivial.

• This proof makes sense. My only question is, how do we know $xA' \neq A'$? In the notation I've been using, this would amount to saying that $\overline{x} \neq \overline{1_A}$, so $x \not \sim 1_A$, but we have $\phi(x) = \phi(1_A) = 1_B$, so that doesn't seem to give a contradiction. Dec 31, 2021 at 23:39
• Because we have $xA'=A'$ if and only if $x\in A'$. If we had $xA'=A'$ then in particular it would mean $x=xe_A\in xA'=A'$, a contradiction. This is why I always like to write cosets as $xA'$, instead of using notations like $\bar{x}$. This makes things much more clear.
– Mark
Dec 31, 2021 at 23:43

Only two facts are required to prove this. The first is the theorem you cited. The second is that $$\ker \pi = A’$$. Note that this means by definition that $$A = \pi^{-1}(0)$$.

A corollary of these facts is that $$\pi$$ is surjective. We will also use this fact.

Note that

$$$$\begin{split} \pi^{-1}(\ker \bar{\phi}) \\ &= \pi^{-1}(\bar{\phi}^{-1}(0)) \\ &= (\bar{\phi} \circ \pi)^{-1}(0) \\ &= \phi^{-1}(0) \\ &= \ker \phi \end{split}$$$$

In the event that $$\bar{\phi}$$ is injective, its kernel is $$0$$. Then $$A’ = \pi^{-1}(0) = \pi^{-1}(\ker \bar{\phi}) = \ker \phi$$ by the above.

Conversely, if $$A’ = \ker \phi$$, then we see that $$\pi^{-1}(0) = A = \ker \phi = \pi^{-1}(\ker \bar{\phi})$$. Recall that $$\pi$$ is a surjective function, and therefore $$\pi^{-1} : P(A / A’) \to P(A)$$ is injective. Then $$\ker \bar{\phi} = 0$$, and thus $$\bar{\phi}$$.