# 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.

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 $$N \cap G_k = G_k$$?

• $N\cap G_i \subset N\cap G_{i+1}$, and $G =\cup_{i=n_0}^{\infty}$ G_iso what can you say if $N\cap G_{n_0} \neq \{1\}$ for some $n_0$? Sep 29, 2013 at 3:56
• I think that what you may have proved is $\;N\cap G_i\lhd G_i\;$ , not $\;N\lhd G_1\;$ as $\;N\;$ could be not contained in some of the $\;G_i\;$ ... Sep 29, 2013 at 3:58

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 $N=N\cap(\cup_{n\ge 1} G_{k_n})=\cup_{n\ge 1}(N\cap G_{k_n})=\cup_{n\ge 1}G_{k_n}=G$.

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

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

• why must such a natural number exist? because if such $k$ goes arbitrarily large, that implies that $G = N$, we are done. Otherwise, by my least favorite well-ordering principle, the maximal $k$ always exists. Sep 29, 2013 at 20:12
• But thank you sosososo much. Your explanation is very clear and very helpful. Sep 29, 2013 at 21:09