$H$ is the subset of $K$, $H$ and also $K$ are the normal subgroups of $(G, * )$. Prove that $H$ is the normal subgroup of $K$, and the quotient group $K/H$ is the normal subgroup of the quotient group $G/H$.

The first part seems to be trivial, or quite easy.

(I am going to use algebraic notation)

My proof is: We know that $H$ is the normal subgroup of $G$, it means that $ {\displaystyle \forall } g ∈G $ and ${\displaystyle \forall } h ∈ H $ : $g^{-1}hg∈ H$

$K$ is the normal subgroup of $G$, it means that $ {\displaystyle \forall } g ∈G $ and ${\displaystyle \forall } k ∈ K $ : $g^{-1}kg∈ K$

And we also know that $H ⊆ K$ whereas $H$ and $K$ are subgroups of $G$, they have to be subset of $G$ it means $H ⊆ K⊆G$.

Than $H$ has to be normal subgroup of $K$ because $ {\displaystyle \forall } g ∈G $ and ${\displaystyle \forall } h ∈ H $ : $g^{-1}hg∈ H$ and this statment is true also for $ {\displaystyle \forall } k ∈K $ and ${\displaystyle \forall } h ∈ H $ : $k^{-1}hk∈ H$ because $K⊆G$.

But what is probelm for me is to continue proof for the second part; quotient group $K/H$ is normal subgroup of quotient group $G/H$.

Thank you very much for any help.

  • $\begingroup$ Have you tried applying the definition? $\endgroup$
    – Kenny Lau
    Dec 28, 2017 at 16:32
  • 1
    $\begingroup$ Nit-pick: I can't see that you have proven that $H$ is a subgroup of $K$. You've proven that it's normal, but not that it's a subgroup. It's a really easy thing to fix, but your proof will be incomplete without it. $\endgroup$
    – Arthur
    Dec 28, 2017 at 16:35
  • $\begingroup$ Yes, you are correct, I have forgotten to make it, it is my mistake. Thank you. $\endgroup$
    – Waney
    Dec 28, 2017 at 16:39

1 Answer 1


The elements of $K/H$ are cosets which we can write as $kH$, and the elements of $G/H$ are cosets which we can write as $gH$. We want to show that $(gH)^{-1}kHgH \in K/H$. By the group law in $G/H$ we have $(gH)^{-1}kHgH = g^{-1}kgH$. Now $g^{-1}kg \in K$ because $K$ is normal in $G$, so $g^{-1}kgH \in K/H$.

  • $\begingroup$ How do we know that $(gH)^{-1}kHgH$ is equal to $g^{-1}kgH$. And what is equal $(gH)^{-1} = ? $ Thank you! $\endgroup$
    – Waney
    Dec 28, 2017 at 17:17
  • $\begingroup$ $gH$ is notation for the coset consisting of $g$ multiplied by any $h \in H$. Since $H$ is normal, $(gH)^{-1} = g^{-1}H$, which we can show as follows: $(gh)^{-1} = h^{-1}g^{-1}$. Since $H$ is normal, $gh^{-1}g^{-1} = h' \in H$, so $h^{-1}g^{-1} = g^{-1}h'$, a member of the coset $g^{-1}H$. Loosely, $H$ "commutes" with all $g \in G$. That's also how we get $g^{-1}HkHgH = g^{-1}kg$. It's this property which enables $G/H$ to be a group. $\endgroup$
    – BallBoy
    Dec 28, 2017 at 17:24

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