Proving Normal Subgroups Let $H$ and $K$ be subgroups of a group $G$, where $H\subseteq K$. If $H$ is a normal as a subgroup (normal subgroup) of $G$, prove that it is also normal as a subgroup (normal subgroup) of $K$?  Definition of a normal subgroup:  A subgroup $K$ of a group $G$ is called a normal subgroup if $k\in K$ and $g\in G$ imply $gkg^{-1}\in K$. If $K$ is a normal subgroup of $G$, we write $K$ as a normal subgroup of $G$.  I need help proving this
 A: You know that $H$ is a normal subgroup of $G$ (noted as $H\trianglelefteq G$) and you want to prove that $H\trianglelefteq K$ given that $H$ and $K$ are subgroups of $G$ and that $H \subseteq K$.
You have to prove that $H$ is a subgroup of $K$ and that it is also normal.
($H$ is a subgroup of $K$) I'll let you prove that.
(Normal) From $H\trianglelefteq G$, you know that for $h\in H$ and $g\in G$, $ghg^{-1}\in H$. 
You want to prove that $H\trianglelefteq K$, so that for $h\in H$ and $k\in K$, $khk^{-1}\in H$. But you know that $K \subseteq G$. What then? (EDIT) Hint : Use the fact that $k\in K$ so $k\in G$.
A: Let us first prove that $H$ is a subgroup of $K$. To do this, I will use the so called subgroup criterion:

If $K$ is a group and $H$ is a non-empty subset. If for all $x, y \in H$ we have that $xy^{-1} \in H$ then $H$ is a subgroup of $K$.

$\textbf{Proof: }$ The prove is quite direct: We know that $H$ and $K$ are subgroups of $G$. Therefore they both contain the neutral element and this shows that $H$ is a nonempty subset of $K$. 
For the second part of this statement, we do not even need that $K$ is a subgroup, but only that $H$ is a subgroup of $G$:
Let $x,y \in H$. Since $H$ is a subgroup of $G$, we have that $H$ is a group itself and hence $xy^{-1} \in H$. 
$\textbf{Remark: }$ Note that in the above proof, we only needed that $K$ was a subgroup to be able to


*

*use the subgroup criterion (only works if the bigger set is a group itself)

*to show that $H$ is a non empty subset of $K$.
For your second part of the proof, let me show that $H$ is also normal in $K$. 
$\textbf{Proof: }$ We have that $H$ is an normal subgroup of $G$, so for every $h \in H$ and for $\color{red}{\text{every } g \in G}$ we have that 
$$ghg^{-1} \in H$$
and since $\color{red}{K \subseteq G}$, it immediately follows that we have for all $k \in K$ that 
$$khk^{-1} \in H$$
which is exactly the condition for $H$ being normal in $K$.
I really hope this edited proof helps you out.
A: Here is my attempt at showing $H$ is a subgroup of $K$ For all $k\in K$ and for all $h\in H$ then there exists $khk^{-1}\in K$ Hence, $H$ is closed under conjugation I am trying to use the conditions of a subgroup like  1) $H$ is nonempty 2) $\forall$ $a,b\in H$ there exists $a,b^{-1}\in H$  to prove $H$ is a subgroup of $K$ 
