Group generated by a set of normal subgroups is normal

For an indexing set $I$, if $\{N_i:i \in I\}$ is a set of normal subgroups of a group $G$, then the smallest subgroup containing all the $N_i$ is given by $\langle N_i:i \in I\rangle = \bigcap_j H_j$, where the $H_j$ are the subgroups of $G$ containing all the $N_i$.

My question is: how do we see that $\bigcap_j H_j = \bigcap_j gH_jg^{-1}$, where $g \in G$ is fixed, as then we can prove that $\langle N_i \rangle$ is normal. I see that $N_i \leq H_j$ implies that $N_i \leq gH_jg^{-1}$ since $N_i$ is normal.

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Shoban gave a direct proof that $\langle N_i : i \in I \rangle$ is normal as long as each $N_i$ is normal. Here is the proof I outlined in the comments that makes use of the intersection definition:

If $N_i \leq H_j$ for all $i$, then $N_i = g^{-1} N_i g \leq g^{-1} H_j g$, so $N_i \leq g^{-1} H_j g$ as well. Since $H_j \mapsto g^{-1} H_j g$ is invertible with inverse $H_j \mapsto g H_j g^{-1}$, the following two sets are equal:

$$\{ H_j : N_i \leq H_j \forall i \} = \{ g^{-1} H_j g : N_i \leq H_j \forall i \}$$

Hence we also have

$$\bigcap\{ H_j : N_i \leq H_j \forall i \} = \bigcap \{ g H_j g^{-1} : N_i \leq H _j \forall i \} = g^{-1}\left( \bigcap \{ H_j : N_i \leq H_j \forall i \} \right) g$$

Hence $\bigcap\{ H_j : N_i \leq H_j \forall i \}$ is normal.

To be clear on the set equality since several people have asked about it: For $X \leq G$, let $I(X) = \{ L : X \leq L \leq G \}$. Check that $$\begin{array}{rl}
&= \{ L : K^g \leq L \leq G \} \\
&= \{ M : K^g \leq M \leq G \} \\
&= \{ L^g : K^g \leq L^g \leq G\} \\
&= \{ L^g : K \leq L \leq G \} \\
&= \{ L^g : L \in I(K) \} \\
&= I(K)^g

The first equality is the definition of $I(K^g)$ with $X=K^g$. The second equality illustrates that the name of the variable does not matter, only that it ranges over all subgroups of $G$ that contain $K^g$. The third equality uses the fact that conjugation is a permutation of the set of subgroups of $G$, so that $\{ L^g : L^g \leq G \} = \{ M : M \leq G \}$ for any $g \in G$. The fourth equality notes that $K \leq L \iff K^g \leq L^g$. The fifth equality is definition of $L \in I(K)$ with $X=K$. The sixth equality is the definition of $Y^g$ for some set $Y$ of subgroups.

Now take $K=N$ to be normal, so that $K=K^g$. Then $I(K) = I(K)^g$ and the set is closed under conjugation.

let $n$ be an element of $\langle N_i \rangle_{i \in I}$ so that it may be written $$n = n_1 n_2 \cdots n_k$$ for some collection of $n_j$ taken from the various $N_i$.

Now consider conjugation $$n^g = (n_1 n_2 \cdots n_k)^g = n_1^g n_2^g \cdots n_k^g$$