What do group theorists mean when they say that $g \mapsto gH$ is undefined when $H$ is non-normal?

Question

If $H$ is a normal subgroup then the function $G \to G/H$ that takes $g$ to $gH$ is well-defined.

If $H$ is an non-normal subgroup then the function $G \to G/H$ that takes $g$ to $gH$ is not well-defined.

Typically when a function $f$ is not well-defined that means that $x = y$ does not imply that $f(x) = f(y)$. So when group theorists say that the morphism $g \mapsto gH$ is not defined for non-normal subgroups, that typically would mean that $g_1 = g_2$ does not imply $g_1H = g_2H$. But it seems obvious from inspection that $g_1 = g_2$ always implies $g_1H = g_2H$, whether $H$ is normal or not.

So what do group theorists mean when they say that the function $g \mapsto gH$ is not well-defined when $H$ is non-normal?

Answer 1

The problem is that it isn't a homomorphism. Take a $g$ s.t. $gHg^{-1} \neq H$ (such $g$ exists, otherwise $H$ is normal). Call $\pi : G \rightarrow G/H$.

If $\pi$ is a homomorphism, then $H = \pi(1) = \pi(gg^{-1}) = \pi(g)\pi(g^{-1})= gHg^{-1} H$.

But $H \neq gHg^{-1}H$! Obviously $H \subseteq gHg^{-1}H$ (if $h \in H$ then $h = g1g^{-1}h \in gHg^{-1}H$). But if $H \supseteq gHg^{-1}H$, then $H \supseteq gHg^{-1}$ which goes against being not normal.

Answer 2

If $H$ is a normal subgroup of $G$, then cosets of $H$ in $G$ form a quotient group in $G$ with the operation defined by $g_1H*g_2H = (g_1g_2)H$. But this is not well-defined if $H$ is not normal, because we might have $g_3\in g_1 H$ and $g_4\in g_2 H$ but $g_1g_2H\ne g_3g_4H$.

Answer 3

Can you find an instance of a group theorist actually saying this?

Because $G/H$, as a set, is perfectly well-defined, and the map from $G$ to this set is a perfectly nice map of sets. It's not of much interest to group theorists as it stands, however, because they tend to care about homomorphisms, and the set $G/H$ lacks a natural group structure unless $H$ happens to be normal in $G$.

It's also the case that a set theorist might define $\sim_H$ to be the equivalence relation on $G$ defined by $a \stackrel{\tiny H}{\sim} b \Leftrightarrow a^{-1}b \in H$, and then write $G/{\stackrel{\tiny H}{\sim}}$ rather than $G/H$, i.e., "mod" in set theory is usually "modulo an equivalence relation" rather than "modulo a subgroup", but that latter expression is really just shorthand for "modulo the natural equivalence relation associated to this subgroup," and it's a lot easier to write $G/H$.