I'm reading through some lecture notes which state:
To show G/K is simple we will show the only normal subgroups of G lying between K and G are K and G
This seems as though I'd need to use an isomorphism theorem, but I can't quite see why it's true.
Simplicity of quotient group
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group-theory
2 Answers
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The theorem you need is, with your notation:
Theorem: there exists a $1-1$ correspondence between the subgroups of the quotient group $\;G/K\;$ and the subgroups of $\;G\;$ containing $\;K\;$ , given by:
$$G/K\ge\overline H\mapsto H:=\left\{\,x\in G\;/\; xK\in\overline H\right\} $$
and
$$K\le H\le G\;,\;\;H\mapsto H/K:=\left\{\,hK\;/\;h\in H\right\}$$
This correspondence maps normal subgroups to corresponding normal subgroups and respects index.
You can google it under "correspondence theorem" or "Fourth Isomorphism Theorem"
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Consider the natural map $q:G\to G/K$. Suppose $G/K$ is not simple, so there is some $1