Prove that the join of any nonempty collection of normal subgroups of a group is an normal subgroup.
Attempt:
Let $\left\langle\bigcup H\right\rangle$ be the join of $H$, where $H$ is the join of any nonempty collection of normal subgroups of $G$.
Let $g\in G$. Then $g\left\langle\bigcup H\right\rangle g^{-1} = g\left\langle \bigcup H_i\right\rangle g^{-1} = \left\langle g\bigcup H_ig^{-1}\right \rangle= \left\langle\bigcup gH_ig^{-1}\right\rangle = \left \langle \bigcup H_i\right\rangle = \left \langle \bigcup H\right\rangle $ So $\left\langle \bigcup H\right\rangle $ is normal.
I don't know if this is a good way to approach this. Can someone please help? Maybe there is a better approach.
Thank you
Let $\{H_a\mid a\in A\}$ be a family of normal subgroups of $G$, we must prove $L=\left\langle\bigcup\limits_{a\in A}H_a\right \rangle \lhd G$. So take an element in $L$. It is of the form $h_1 h_2\dots h_n$ with $h_i\in H_a$ for some $a\in A$ for every $1\leq i\leq n$.
We must prove $g(h_1 h_2\dots h_n)g^{-1}\in L$
Now notice $g(h_1 h_2\dots h_n)g^{-1}=(gh_1g^{-1})(gh_2g^{-1})\dots (gh_ng^{-1})$. And because each $H_a$ is normal we notice if $h_i\in H_a$ then $gh_ig^{-1}\in H_a$. Therefore $g(h_1 h_2\dots h_n)g^{-1}\in L$ as desired.