anon anon

2022-06-13

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

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