For a group $G$, the normal core $H_G$ of a subgroup $H \le G$ is defined as the intersection of the conjugates of $H$, i.e., $$H_{G} = \bigcap_{a \in G} a H a^{-1}.$$ It is remarked that $H_{G}$ is the largest normal subgroup of $G$ that is contained in $H$.
I am able to check that $H_G$ is indeed a normal subgroup of $G$ but fail to show its maximality. That is,
we want to prove that if $K \le H, K \triangleleft G$, then $K \subseteq H_G$.
Here is my trial:
Given $K \le H, K \triangleleft G$. Assume that, by contradiction, $K \nsubseteq H_G$.
$$K \nsubseteq H_G \implies K \nsubseteq \bigcap_{a \in G} aHa^{-1} \implies \exists a \in G (\exists k \in K, k \notin aHa^{-1}).$$
By normality of $K$ in $G$, we have $aka^{-1} \in K$. On the other hand, we have $k \in K \subseteq H \implies aka^{-1} \in aHa^{-1}$. Therefore, $aka^{-1} \in H \cap aHa^{-1}$.
Similarly, $a^{2}k(a^{-1})^{2} \in H \cap aHa^{-1}, a^{3}k(a^{-1})^{3} \in H \cap aHa^{-1}$, and so on.
If the order of $a$ is finite (i.e., $\text{ord}(a) = n$), we would obtain $(eke = k)\in H \cap aHa^{-1}$, contrary to the assumption $k \notin aHa^{-1}$.
As you see, my argument relies on the assumption of finiteness of $\text{ord}(a)$ which is not necessarily true. Hence, my problem is as follows:
- Can my argument be fixed?
- If my argument is not on the right track, could you offer me a hint of the proof of the maximality of normal core?
We can fix it by noting that
$$k \notin aHa^{-1} \implies a^{-1}ka \notin a^{-1}aHa^{-1}a = H,$$
but since $K$ is normal, we have $a^{-1}ka \in K \subset H$, reaching a contradiction.
Another way to put the argument is that by the normality of $K$, we have $$a^{-1}Ka \subset K$$ for all $a \in G$, but $$a^{-1}Ka \subset H \iff K \subset aHa^{-1},$$
so
$$\bigl(\forall a\in G\bigr) \bigl( K \subset aHa^{-1}\bigr) \iff K \subset \bigcap_{a\in G} aHa^{-1}.$$