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.


2 Answers 2


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.

  • $\begingroup$ 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. $\endgroup$
    – Brad G.
    Dec 31, 2021 at 23:39
  • 1
    $\begingroup$ 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. $\endgroup$
    – 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

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

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 $\bar{\phi}$.


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