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Let $\{N_i\}_{i\in I}$ be a collection of normal subgroups of $\mathbf{G}$, then show that $$\left\langle \bigcup_{i \in I} N_i \right\rangle \trianglelefteq \mathbf{G}$$

We proved this in the following way:

Let $g\in \mathbf{G} , x\in \bigcup_{i \in I} N_i \therefore x\in N_{i_0}, \exists i_0 \in I$

$\therefore gxg^{-1} \in N_{i_0} \subseteq \bigcup_{i \in I} N_i ,since \quad N_{i_0}\trianglelefteq \mathbf G$

$\therefore g(\bigcup_{i \in I} N_i)g^{-1}= \bigcup_{i \in I} N_i $

$\implies \left\langle g(\bigcup_{i \in I} N_i)g^{-1}\right\rangle = \left\langle \bigcup_{i \in I} N_i\right\rangle (*) $

$\left\langle g(\bigcup_{i \in I} N_i)g^{-1}\right\rangle = g\left\langle \bigcup_{i \in I} N_i\right\rangle g^{-1} (**) $

$(**)$ follows by a theorem we showed in class

From $(*)$ and $(**)$ , $$\left\langle \bigcup_{i \in I} N_i \right\rangle \trianglelefteq \mathbf{G}$$ Now, I didn't understand how in $(*)$ we showed there is an equality. How that one followed? Could someone explain it?

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Recall that a subgroup $N$ of $G$ is normal if for every $g\in G$ we have $gNg^{-1}=N$.

In particular $$g\left(\bigcup_{i\in I} N_i\right) g^{-1} = \bigcup_{i\in I} gN_i g^{-1} = \bigcup_{i\in I} N_i$$

The first equality is a simple equality between sets, the second equality follows from the fact that $N_i$ is normal in $G$.

This clearly implies that $\left<g\left(\bigcup_{i\in I} N_i\right) g^{-1}\right> = \left<\bigcup_{i\in I} N_i\right>$.

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  • $\begingroup$ Isn't the definition of normal subgroups this one: " a subgroup $N$ of $G$ is normal if for every $g\in G$ we have $gNg^{-1}\subseteq N$ ? @yanko $\endgroup$ Oct 21 '17 at 13:21
  • $\begingroup$ @LeylaAlkan these two definitions are equivalent, see en.wikipedia.org/wiki/Normal_subgroup $\endgroup$
    – Yanko
    Oct 21 '17 at 13:24
  • $\begingroup$ They're equivalent because $gNg^{-1}\subseteq N$ if and only if $N\subseteq g^{-1}Ng$. $\endgroup$
    – Yanko
    Oct 21 '17 at 13:25

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