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Every finite group has a composition series.

The proof of this statement is as follows

Proof. If $|G| = 1$ or $G$ is simple, then the result is trivial. Suppose $G$ is not simple, and the result holds for all groups of order $< |G|$. Let $H$ be a maximal normal subgroup of $G$. By the induction hypothesis $H$ has a composition series ${e} \subset H_1 \subset \cdots \subset H$. Therefore, $G$ has a composition series ${e} \subset H_1 \subset \cdots \subset H \subset G$.

My question is, how can it be guaranteed that there exists a maximal normal subgroup of $G$ of order $< G$?

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    $\begingroup$ The collection of subgroups is a finite set, so you can just pick the maximal normal one. $\endgroup$ Jun 15 '14 at 10:45
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My question is, how can it be guaranteed that there exists a maximal normal subgroup of $G$ of order $< G$?

Another argument: If the set $S$ of normal subgroups of $G$ is $\emptyset$ then $G$ is simple and there is the composition series $1 \subset G $.

Otherwise every chain in $S$ ordered with respect to inclusion has an upper-bound, so we can apply Zorn lemma and there is a maximal element.

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