# Quotient group implies normality

We know that for a group $$G$$ and a normal subgroup $$H \triangleleft G$$, the operation $$g_1Hg_2 H := g_1g_2H$$ is well-defined and in fact results in a group structure on $$G/H$$. Conversely, I want to show that if this operation is well-defined, my subgroup will in fact be normal.

To this end, I let $$g_1, g_2 \in H$$ and for some $$h \in H$$, $$g_1' := g_1h, g_2' := g_2h$$. Then we have

\begin{align} g_1g_2 H &= g_1 H g_2 H \\ &= g_1hH g_2hH \\ &= g_1'H g_2'H \\ &= g_1'g_2'H \\ &= g_1hg_2hH \\ &= g_1hg_2H. \end{align}

Hence there must exist some $$\tilde{h} \in H$$ such that $$g_1g_2 = g_1hg_2\tilde{h}$$, so $$g_2 = hg_2\tilde{h}$$, or $$g_2\tilde{h}^{-1} = hg_2$$.

Now if $$g_2\tilde{h}$$ was an arbitrary element of $$g_2H$$, we'd be done. However, since we've only shown that such an $$\tilde{h} \in H$$ exists, we're not.

Hence let $$x \in g_2H$$. Then there exists some $$y \in H$$ such that $$x = g_2y$$. What I want to show now is that there must also exist a $$z \in H$$ such that $$x = zg_2$$. I've tried this, but I didn't get any further:

\begin{align} x &= g_2y \\ &= g_2\left(\tilde{h}^{-1}\tilde{h}\right)y \\ &= hg_2 \tilde{h}y. \end{align}

Can anyone find me my $$z$$?

• See this question. – Dietrich Burde Feb 24 '19 at 17:47
• What is your definition of normal? (In fact, some take "being a kernel" as definition, others first show that several properties are equivalent and thereby motivate the definition of the concept of normality ...) – Hagen von Eitzen Feb 24 '19 at 17:48
• Ah yes, my definition of normal is that for all $g \in G$, $gH = Hg$. That's why I'm trying to show that $x = zg_2$ for some $z \in H$. – Jos van Nieuwman Feb 24 '19 at 17:53

The definition $$gH=Hg$$ for all $$g\in G$$ is equivalent to $$ghg^{-1}\in H$$ for all $$g\in G,h\in H$$. So if $$H$$ is not normal then there are $$g\in G$$ and $$h\in H$$ such that $$ghg^{-1}\notin H$$. But that means $$gHg^{-1}H\ne gg^{-1}H=H$$. And it is obvious $$gHg^{-1}H$$ can't be equal to any coset which is not $$H$$. So we have to conclude it is not a coset at all.