For all $g\in G$, is it true that $gHg^{-1} \subseteq H $ if H is a normal subgroup of G.

Question

Assume that $H$ is a normal subgroup of group $G$.

Is it true that for all $g∈G$ one has $gHg^{-1} \subseteq H$ ?

I know that if H is a normal subgroup of G, then $ghg^{-1}$∈ $H$ where $h$∈$H$ and $g,g^{-1}∈G$. I am just unsure how this slight change in statement (i.e. using the entire group $H$ instead of just an element of $H$) affects the situation?

Answer 1

Write down explicitly:

$$\forall\,g\in G\;,\;\;gHg^{-1}:=\{\,ghg^{-1}\;:\;h\in H\,\}$$

Now, since (your definition) $\;H\lhd G\implies ghg^{-1}\in H\;,\;\;\forall\,h\in H$ , and thus indeed $\;gHg^{-1}\subset H\;$ since every element in the definition written above is in $\;H\;$ .

Answer 2

First, the statement $g H g^{-1} \in H$ (which you had originally written, and then edited) is meaningless: here $g H g^{-1}$ is the set $\{g h g^{-1} : h\in H\}$ of elements of the form $g h g^{-1}$ for $h\in H$, and it doesn't make sense to ask whether it belongs to $H$. (Well, in set theory, it does make sense, because everything in mathematics is a set, but it's like asking whether $42 \in \sqrt{2}$, not a "reasonable" thing to write.)

Since $g H g^{-1} = \{g h g^{-1} : h\in H\}$, writing $g H g^{-1} \subseteq H$ is exactly equivalent to writing $g h g^{-1} \in H$ for every $h \in H$. So if that's your definition of a normal subgroup, yes, saying that $g H g^{-1} \subseteq H$ for all $g \in G$ is exactly equivalent to saying that $g h g^{-1} \in H$ for all $g\in G$ and $h \in H$.

Another equivalent property is that $g H g^{-1} = H$ for all $g\in G$. It trivially implies $g H g^{-1} \subseteq H$ for all $g\in G$, but in fact it is equivalent because the latter (applied to $g^{-1}$) also gives $g^{-1} H g \subseteq H$, which upon multiplication by $g$ on the left and $g^{-1}$ on the right, implies $H \subseteq g H g^{-1}$, so in fact an equality.