Problem

Prove part (1) of the Jordan-Hölder Theorem by induction on \(|G|\).

Solution

Suppose \(|G| = p\) for some prime \(p\).

If there is a subgroup \(H\), by Lagrange’s Theorem, \(|H|\) divides \(|G|\), so \(H = G\) or \(H = 1\). Therefore, \(1\) and \(G\) are the only normal subgroups of \(G\). Since \(|G| = p > 1\), \(G\) is simple. Thus, \(1 \le G\) is a composition series of \(G\) because \(1 \trianglelefteq G\) and \(G/1 = G\) is simple.

Note this establishes the base case for the induction, covering \(|G| = 2\) and \(|G| = 3\). Now consider \(|G| = k\) for some \(k > 3\) while assuming that the statement holds for every group of order smaller than \(k\).

Suppose \(G\) is simple. Then \(1 \le G\) is a composition series of \(G\) because \(1 \trianglelefteq G\) and \(G/1 = G\) is simple.

Suppose \(G\) is not simple. Then there exists at least one normal subgroup of \(G\) that is not \(1\) or \(G\). Since \(G\) is finite, among those normal subgroups, there is the maximal proper normal subgroup \(H\) such that all the following hold.

• \(1 \ne H\) and \(H \ne G\)
• \(H \trianglelefteq G\)
• if \(M \trianglelefteq G\) and \(H \le M\), then \(M = H\) or \(M = G\)

Since \(|H| < |G|\), by our induction assumption, \(H\) has a composition series, say,

$$ 1 = H_0 \le H_1 \le … \le H_{n-1} \le H_n = H. $$

for some positive integer \(n\). We argue that \(G\) has the following composition series.

$$ 1 = H_0 \le H_1 \le … \le H_{n-1} \le H \le H_{n+1} = G $$

Given the composition series for \(H\), it suffices to show that

• \(H \trianglelefteq G\), and
• \(G/H\) is simple.

\(H \trianglelefteq G\) holds by definition of \(H\). Since \(H\) is a proper subgroup of \(G\), using Lagrange’s Theorem, \(|G/H| = \frac{|G|}{|H|} > 1 \). It remains to show that \(1\) and \(G/H\) are the only normal subgroups of \(G/H\).

Suppose for a contradiction that there is a normal subgroup \(K/H\) of \(G/H\) such that \(K/H \ne 1\) and \(K/H \ne G/H\). Then, by the Fourth Isomorphism Theorem,

• \(H \le K\),
• \(1 \trianglelefteq K \trianglelefteq G\),
• \(1 \ne K\), and \(K \ne G\).

Because \(H \le K\) and \(H\) is the maximal proper normal subgroup, \(K = H\) or \(K = G\). That would result in \(K/H = H/H = 1\) or \(K/H = G/H\), contradicting our assumption that \(K/H \ne 1\) and \(K/H \ne G/H\). Therefore, there’s no such normal subgroup \(K/H\) of \(G/H\).

Hence, \(G/H\) is simple. This concludes the proof by induction.

$$\tag*{$\blacksquare$}$$