If $G_1 \leq G_2 \leq \cdots \leq G$ such that $G = \bigcup_{i=1}^\infty G_i$ and each $G_i$ is simple then $G$ is simple.

Question

Prove that if there exists a chain of subgroups $G_1 \leq G_2 \leq \cdots \leq G$ such that $G = \bigcup_{i=1}^\infty G_i$ and each $G_i$ is simple then $G$ is simple.

I have proved the following lemma:

Let $N \triangleleft G$. Then $N\cap G_i \triangleleft G_i$ for each $G_i$.

$N \cap (\cup_{i=1}^\infty G_i) = \cup_{i=1}^\infty (N \cap G_i)$

Since each $G_i$ is simple, we have $N \cap G_i \in \{ 1, G_i \}$.

Suppose $N \cap G_i = 1$ for all $G_i$. Then $$N = N \cap G = N \cap (\bigcup_{i=1}^\infty G_i) = \bigcup_{i=1}^\infty (N \cap G_i) = \bigcup_{i=1}^\infty 1 = 1.$$ Hence $N = 1$.

Then I am not quite sure how to handle the case that $N \cap G_k = G_k$?

Answer 1

If there is $k\ge 1$ such that $N\cap G_k=G_k$, then you have two cases:

1) There are finitely many such $k$, and then there exists an $n_0\ge 1$ such that $N\cap G_n=\{1\}$ for all $n\ge n_0$, so $N=N\cap(\cup_{n\ge n_0}G_n)=\cup_{n\ge n_0}(N\cap G_n)=\{1\}$. (Here I've used that $\cup_{n\ge n_0}G_n=G$.)

2) There are infinitely many such $k$, and then there exists a sequence $(k_n)_{n\ge 1}$ such that $N\cap G_{k_n}=G_{k_n}$ for all $n\ge 1$, so $N=N\cap(\cup_{n\ge 1} G_{k_n})=\cup_{n\ge 1}(N\cap G_{k_n})=\cup_{n\ge 1}G_{k_n}=G$.

Answer 2

For any two sets $\;A,B\;$ , remember that $\;A\cap B=A\iff A\subset B\;$ , so:

If there exists

$$n\in\Bbb N\;\;s.t.\;\;N\cap G_n=G_n\iff G_n\le N$$

Let now $\;k\in\Bbb N\; $ be the maximal subindex s.t. $\;G_k\le N\;$ (why must such a natural number exist?), so $\;G_{k+1}\rlap{\;\,/}\le N\;$ and thus $\;N\cap G_{k+1}=1\;$ . Obviously $\;k\ge n\;$ .

But then

$$G_n=N\cap G_n\le N\cap G_{n+1}=1\;,\;\;\text{which is absurd}\ldots$$