Show that if \(N\) is a simple normal subgroup of a group \(G\) and \(G/N\) has a composition series, then \(G\) has a composition series.
Solution
Let \[ G/N=\overline G_0\trianglerighteq\overline G_1 \trianglerighteq\cdots\trianglerighteq\overline G_r=\{N\} \] be a composition series for \(G/N\), and let \[ \pi:G\longrightarrow G/N \] be the quotient map. Put \(G_i=\pi^{-1}(\overline G_i)\).
1. The preimages form a normal series\[ G=G_0\trianglerighteq G_1\trianglerighteq\cdots \trianglerighteq G_r=N. \]
Since \(\overline G_{i+1}\trianglelefteq\overline G_i\), the preimage \(G_{i+1}\) is normal in \(G_i\). Also \[ G_0=\pi^{-1}(G/N)=G, \qquad G_r=\pi^{-1}(\{N\})=N. \]
2. Every factor \(G_i/G_{i+1}\) is simple.
The restriction of \(\pi\) induces an isomorphism \[ G_i/G_{i+1}\cong \overline G_i/\overline G_{i+1} \] by the third isomorphism theorem, since \(\overline G_i=G_i/N\) and \(\overline G_{i+1}=G_{i+1}/N\). The quotient on the right is simple because the given series of \(G/N\) is a composition series.
3. Appending the term \(\{e\}\) gives a composition series\[ G=G_0\trianglerighteq\cdots\trianglerighteq G_r=N \trianglerighteq\{e\}. \]
By 2 all factors above \(N\) are simple. The final factor \[ N/\{e\}\cong N \] is simple by hypothesis. Hence every factor in the displayed normal series is simple, which is precisely a composition series for \(G\).