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Suppose $G_1$, $G_2$ are groups and $N_1 \subset G_1$, $N_2 \subset G_2$ are normal subgroups. Let $f:G_1\to G_2$ be a homomorphism. When is $\bar{f}(N_1 g) = N_2 f(g)$ a well-defined homomorphism from $G_1/N_1$ to $G_2/N_2$?

user26857
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Anonnon
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1 Answers1

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It isn't necessarily well-defined. A necessary condition is that, for $g\in N_1$, $N_2f(g)=N_2$, that is, $f(g)\in N_2$. In other words, we need that $f(N_1)\subset N_2$.

This condition is also sufficient, because in this case $$ N_1\subset\ker \pi\circ f $$ where $\pi\colon G_2\to G_2/N_2$ is the canonical projection.

A general result is the following: if $\varphi\colon G\to G'$ is a group homomorphism and $N$ is a normal subgroup of $G$ such that $N\subset\ker\varphi$, then $$ Ng\mapsto \varphi(g) $$ is a well defined group homomorphism $G/N\to G'$. It is well defined because, if $Ng_1=Ng_2$, then $g_1g_2^{-1}\in N$, so $g_1g_2^{-1}\in \ker\varphi$ and therefore $\varphi(g_1)=\varphi(g_2)$. It is a homomorphism in an obvious way.

A counterexample to the general assertion is easy to make. Consider the identity $1_{\mathbb{Z}}\colon\mathbb{Z}\to\mathbb{Z}$; let $N_2=\{0\}$ and $N_1=2\mathbb{Z}$.

egreg
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