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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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How do the sets $\{ H_j : N_i \leq H_j \}$ and $\{ gH_jg^{-1} : N_i \leq H_j \}$ compare? – Jack Schmidt Apr 23 '13 at 14:09
@Jack I see that they are not equal since $H_j \not= gH_jg^{-1}$ in general? – user50229 Apr 23 '13 at 14:39
While $H_j \neq gH_jg^{-1}$ in general, the two sets I mention are equal. – Jack Schmidt Apr 23 '13 at 14:52
I think my suggestion is very elementary. Take an element in the left subset, prove it is in the right, then vice versa. If two sets are equal, then so are their intersections. – Jack Schmidt Apr 23 '13 at 16:34
Is this correct for one direction: $N_i \leq H_j$, so $g N_i g^{-1} \leq g H_j g^{-1}$. Since $N_i$ is normal, $N_i \leq g H_j g^{-1}$. This shows $g H_j g^{-1} \in \{H_j\}$, so $\{g H_j g^{-1} \} \subseteq \{H_j\}$. – user50229 Apr 23 '13 at 18:44
up vote 2 down vote accepted

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 $$\begin{array}{rl} I(K^g) &= \{ L : K^g \leq L \leq G \} \\ &= \{ M : K^g \leq M \leq G \} \\ &= \{ L^g : K^g \leq L^g \leq G\} \\ &= \{ L^g : K \leq L \leq G \} \\ &= \{ L^g : L \in I(K) \} \\ &= I(K)^g \end{array}$$

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.

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the bijection function doesn't mean that the sets are equal . ,e.g. let $Z$ denotes the integers ,let $f:Z\rightarrow ${$2k:k\in Z$} ,$f(x)=2x$ ,is a bijection but the domain and the range are clearly are not equal sets. so if the number of the $H_i$'s is infinite ,then it may that the sets are not equal , so this argument doesn't work for any infinite abelian group with infinite number of subgroups like the integers under addition . your argument works perfectly when the $G$ is finite or\and the number of the normal subgroups is finite. – Maths Lover Jun 21 '13 at 0:00
@MathsLover: I tried to explain. The ”bijection” is a permutation, or just a way of changing the name of a variable. – Jack Schmidt Jun 21 '13 at 15:33
thank you very much for clarification :) . now it completely make sense . Regards , @Jack Schmidt . – Maths Lover Jun 21 '13 at 20:41

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

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