Hello so I need to proof that Center is subgroup of normalizer which I did but I did it using the following definition of $N_G(A) = \{h \in G : hA = Ah\}$.
But the definition that I was given is $N_G(A) = \{ h \in G : hah^{-1} \in A \;\forall a \in A\}$.
I can see that the following conditions are equivalent intuitively but I want to prove it.
Unfortunately, the two definitions are not equivalent if the groups are infinite. Specifically, the first definition is the standard definition of normalizer, and the second definition is "wrong".
As Matthew details in his answer, the second condition gives us $hA\subseteq Ah$. Furthermore, we know that the function $Ah \rightarrow hA$ defined by conjugation by $h$ is injective, and thus its image has the same cardinality as the domain. From the above two facts we can conclude that $hA$ and $Ah$ have the same cardinality.
If $A$ is finite that indeed implies that $hA=Ah$ since one is contained in the other and they have the same finite cardinality.
If $A$ is infinite, however, the above argument breaks down. Indeed, as shown here, it is possible that $hA\subsetneq Ah$. Moreover, as shown here, the second condition does not even necessarily define a group!
What you want to show is that $$ hA = Ah \iff \forall a \in A,\ h a h^{-1} \in A $$
Suppose $hA = Ah$, and let $a \in A$. Since $hA = Ah$, and $ha \in hA$, we have $ha = a'h$ for some $a'\in A$. Then $$ ha = a'h \implies hah^{-1} =a', $$ evidently in $A$.
On the other hand, suppose $ h a h^{-1} \in A$ for all $a \in A$. Let $x \in hA$. Then $x = ha$ for some $a \in A$. Since $hah^{-1} \in A$, we must have $$ hah^{-1} = a' \implies x = ha = a'h $$ So $x \in Ah$ too. This shows $hA \subseteq Ah$. We can similarly show $Ah \subseteq hA$; hence $hA = Ah$.
