To prove that if $K\u25c1H$ (K is a normal subgroup of H), then $K\u25c1G$ (K is a normal subgroup of G), we need to show that K satisfies the conditions of a normal subgroup with respect to G.

Recall that a subgroup K is considered normal in a group G if and only if for every element g in G, the conjugate of K by g, denoted by $gK{g}^{-1}$, is contained in K.

Let's prove this statement formally. Assume K is a normal subgroup of H, meaning for every h in H, $hK{h}^{-1}\subseteq K$.

Now, let's take an arbitrary element g in G. Since H is a subgroup of G (as K is a subgroup of H), g is also in H.

Consider the conjugate of K by g in G:

$gK{g}^{-1}$

Since g is in H, we can express g as h for some h in H. Therefore:

$gK{g}^{-1}=hK{h}^{-1}$

Since K is a normal subgroup of H, we know that for every h in H, $hK{h}^{-1}\subseteq K$. Thus, we have:

$gK{g}^{-1}=hK{h}^{-1}\subseteq K$

Therefore, we can conclude that K is a normal subgroup of G, as for every element g in G, the conjugate of K by g, $gK{g}^{-1}$, is contained in K.

To prove that if $K\u25c1H$, then $K\u25c1G$, we need to show that K satisfies the conditions of a normal subgroup with respect to G.

Assume $K\u25c1H$, meaning for every $h\in H$, $hK{h}^{-1}\subseteq K$.

Now, let's take an arbitrary element $g\in G$. Since $H$ is a subgroup of G (as $K$ is a subgroup of $H$), $g$ is also in $H$.

Consider the conjugate of K by g in G:

$gK{g}^{-1}$

Since $g$ is in $H$, we can express $g$ as $h$ for some $h\in H$. Therefore:

$gK{g}^{-1}=hK{h}^{-1}$

Since K is a normal subgroup of H, we know that for every $h\in H$, $hK{h}^{-1}\subseteq K$. Thus, we have:

$gK{g}^{-1}=hK{h}^{-1}\subseteq K$

Therefore, we can conclude that $K$ is a normal subgroup of $G$, as for every element $g\in G$, the conjugate of $K$ by $g$, $gK{g}^{-1}$, is contained in $K$.