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 $z$?

  • $\begingroup$ See this question. $\endgroup$ Feb 24, 2019 at 17:47
  • $\begingroup$ 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 ...) $\endgroup$ Feb 24, 2019 at 17:48
  • $\begingroup$ 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 $z \in H$. $\endgroup$ Feb 24, 2019 at 17:53

1 Answer 1


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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.