Proving that infinite union of simple groups is also simple group

Question

Given that: $$G_1\subseteq G_2\subseteq G_3\subseteq \dots\subseteq G_n \subseteq G_{n+1}\subseteq \cdots$$ are all simple groups. Prove that $$G=\bigcup_{n=1}^{\infty}G_n$$ is also a simple group.

So first of all, I proved that $G$ is a group, there wasn't too much work to do there, but how can I prove that $G$ is simple?

thank you in advance.

Answer 1

Hint:

$$\;1\lneq N\lhd G\;,\;\;N\lneq G\;\implies \exists\;k\in\Bbb N\;\;s.t.\;\;G_k\rlap{\;\,/}\subset N $$

Well, take a peek at $\;G_k\cap N\lhd G_k\;$ . By the above, it must be $\;G_k\cap N=1\;$ and etc.

Answer 2

A group $H$ is simple if, and only if, every non-trivial group homomorphism $H\to H'$, into any group $H'$, is injective.

So, let $\psi :G\to H'$ be a non-injective homomorphism, and suppose that $\psi (g_1)=\psi (g_2)$, with $g_1\ne g_2$. By the construction of $G$, we have that $g_1,g_2\in G_k$ for all $k>k_0$, and that $\psi \mid _{G_k}:G_k\to H'$ is a homomorphism, where $k_0$ is some natural number. But then since $G_k$ is simple it follows that $\psi$ is trivial on $G_k$. This is true for all $h>k_0$, and so $\psi $ is trivial on $G$.