Let $N_1\leq N_2\leq N_3\leq\cdots$ be normal subgroups of $G$. Show that $\bigcup_{i=1}^\infty N_i$ is a normal subgroup of $G$.

Question

Let $N_1 \leq N_2\leq N_3\leq \cdots$ be normal subgroups of $G$. Show that $\bigcup_{i=1}^{\infty}N_i$ is a normal subgroup of $G$.

Please help me to verify my attempt, in case I have left any details.

My attempt: Since $N_1 \leq N_2\leq N_3\leq \cdots$, we have that $N_1 \subseteq N_2\subseteq N_3\subseteq \cdots$, so $\bigcup_{i=1}^{\infty}N_i$ is a subgroup of $G$. (Justification)

Now if $a\in N_i$ and $g\in G$, then $g^{-1}ag\in N_i$. Thus $g^{-1}ag\in \bigcup_{i=1}^{\infty} N_i$ and we are done.

Answer 1

It seems fine to me, but your "justification" is a little off (as it only proves the result for two subgroups); try this instead.

Oh, and you've overused $i$ in your last paragraph. You can remedy this as follows. Suppose $a\in H:=\bigcup_{i=1}^\infty N_i$. Then $a\in N_j$ for some $j$. Since $N_j$ is normal, $gag^{-1}\in N_j$ for all $g\in G$. But then $gag^{-1}\in H$. Hence $H\unlhd G$.