Theorem 21
If \(G\) is a finite abelian group and \(p\) is a prime dividing \(|G|\), then \(G\) contains an element of order \(p\).
Proof
The proof is by induction where the induction step states that the statement holds for \(G\) of order \(k\) if it holds for every \(G\) of order smaller than \(k\).
\(p \mid |G|\) and thus \(|G| > 1\). Therefore, there exists \(x \in G\) such that \(x \ne 1\).
Consider the case where the \(G\) is of the smallest possible order, where \(|G| = p\). By Lagrange’s Theorem, \(|x| = p\) because \(\langle x \rangle\) is a non-trivial subgroup of \(G\) and its order must divide \(|G|\). Therefore, \(G\) contains an element, namely \(x\), of order \(p\).
Next, consider cases where \(p < |G|\).
Suppose \(p \mid |x|\), and \(|x| = pm\) for some positive integer \(m\). Then, by Proposition 2.5(3),
$$ |x^m| = \frac{pm}{m} = p. $$
Therefore, \(G\) contains an element \(x^m\) of order \(p\).
Suppose \(p \nmid |x|\). Let \(N = \langle x \rangle\). Since \(G\) is abelian, for all \(g \in G\), for all \(a \in N \le G\), \(gag^{-1} = gg^{-1}a = a \in N\). Therefore, \(N \trianglelefteq G\) by Theorem 6. By Lagrange’s Theorem, \(|G/N| = \frac{|G|}{|N|}\). Since \(x \ne 1\), \(|N| > 1\), and thus \(|G/N| < |G|\). Also, \(p \nmid |N|\) but \(p \mid |G|\), so \(p \mid |G/N|\).
That is, \(G/N\) is a finite abelian group of order smaller than \(|G|\) and divisible by \(p\). Thus, by our induction assumption, we can assume that there exists \(yN \in G/N\) of order \(p\).
Because \(|yN| = p\), \(yN\) is not the identity. Therefore, \(yN \ne N\) and thus \(y \notin N\). Additionally, \((yN)^{p} = y^pN = 1\), so \(y^p \in N\). Therefore, \(\langle y^p \rangle \ne \langle y \rangle\), \(|y^p| < |y|\), and \(\langle y^p \rangle \le \langle y \rangle \le G\).
By Proposition 2.5 (2), \(|y^p| = \frac{|y|}{\gcd(|y|, p)}\). Since \(|y| \ne |y^p|\), \(\gcd(|y|, p) \ne 1\), and thus \(p \mid \gcd(|y|, p)\). Therefore, \(p \mid |y|\) as well. That is, \(|y| = pm\) for some positive integer \(m\). By Proposition 2.5 (3), \(|y^m| = \frac{pm}{m} = p\).
Hence, \(G\) contains an element \(y^m\) of order \(p\).
$$\tag*{$\blacksquare$}$$
Simple group
A (finite or infinite) group \(G\) is called simple if \(|G| > 1\) and the only normal subgroups of \(G\) are \(1\) and \(G\).
Composition series
In a group \(G\) a sequence of subgroups
$$ 1 = N_0 \le N_1 \le N_2 \le … \le N_{k-1} \le N_k = G $$
is called a composition series if \(N_i \trianglelefteq N_{i+1}\) and \(N_{i+1}/N_i\) is a simple group, \(0 \le i \le k-1\). If the above sequence is a composition series, the quotient groups \(N_{i+1}/N_i\) are called composition factors of \(G\).
Theorem 22. Jordan-Hölder
Let \(G\) be a finite group with \(G \ne 1\). Then
- \(G\) has a composition series and
- The composition factors in a composition series are unique, namely, if \(1 = N_0 \le N_1 \le … \le N_r = G\) and \(1 = M_0 \le M_1 \le … \le M_s = G\) are two composition series for \(G\), then \(r = s\) and there is some permutation, \(\pi\), of \(\{1,2,…,r\}\) such that
$$ M_{\pi(i)}/M_{\pi(i)-1} \cong N_i/N_{i-1},\quad 1 \le i \le r. $$
Proof
See 3.4.06.