Let $\phi : G \to H/N$ be a homomorphism where $G$ and $H$ are groups and let $M \unlhd G$ and $N \unlhd H$.

Now when does $\phi$ induces a homomorphism $\phi^*$ from $G/M \to H/N\ ?$

  1. When $M \subseteq \text{ker}(\phi)$
  2. When $\text{ker}(\phi) \subseteq M$.
  3. or in both cases?

In the both cases, induced homomorphism looks like $$\phi^*(g+M)=\phi(g)+N$$

So is it a homomorphism in both cases? If yes, Why does it has to to satisfy a containment relation with $ker(\phi)$. What if $M$ is just some random subgroup of $G$?

  • $\begingroup$ Observe that $\phi(g)$ is, according to your definition of $\phi$, already an element of $H/N$, i.e., it is a coset of $N$. It doesn't really do anything / make sense to write $\phi(g)+N$. $\endgroup$ Commented Jun 16, 2015 at 22:25

1 Answer 1


Only when $M\subseteq \ker(\phi)$ (and of course, we also need $M$ to be normal in $G$, which is necessary for $G/M$ to be a group in the first place.)

First, forget about $H$ and $N$, which play no role separately in your question. We may as well just set $Q=H/N$. Thus, you have a homomorphism $\phi:G\to Q$, and want to know when there is an induced homomorphism $\phi^*:G/M\to Q$.

In order for the (proposed) formula $$\phi^*(gM)=\phi(g)$$ to make a well-defined function $\phi^*:G/M\to Q$, you need to have $$g_1M=g_2M\implies \phi(g_1)=\phi(g_2)$$ which by the first isomorphism theorem is equivalent to $$g_1M=g_2M\implies g_1\ker(\phi)=g_2\ker(\phi)$$ which is equivalent to $$g\in M\implies g\in\ker(\phi)$$ which is equivalent to $M\subseteq\ker(\phi)$.

  • $\begingroup$ Is there any way to induce some homomorphism if $\ker(\phi)\subseteq M?$ $\endgroup$
    – blabla
    Commented Jun 16, 2015 at 22:26
  • 1
    $\begingroup$ No. If you can write $\phi:G\to Q$ as a composition of the quotient map $r:G\to G/M$ followed by a homomorphism $\phi^*:G/M\to Q$, then the image of $\phi^*$ must the same as the image of $\phi$. But if $\ker(\phi)\subsetneq M$, then $\lvert G/M\rvert$ is smaller than $\lvert G/\ker(\phi)\rvert=\lvert \phi(G)\rvert$, and therefore there couldn't possibly be any such $\phi^*$. (This shows it's not possible for finite groups $G$, which demonstrates that it couldn't be a general fact.) $\endgroup$ Commented Jun 16, 2015 at 22:30

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