Problem

Prove that if \(G\) is an abelian simple group, then \(G \cong Z_p\) for some prime \(p\). Do not assume \(G\) is a finite group.

Solution

Suppose \(G\) is an abelian simple group.

Suppose for a contradiction that there exists \(H \le G\) such that \(H \ne 1\) and \(H \ne G\). Since \(G\) is simple, for some \(h \in H\) and \(g \in G\), \(ghg^{-1} \notin H\). Since \(G\) is abelian, it follows that \(gg^{-1}h = h \notin H\), leading to a contradiction. Therefore, no such \(H\) exists. That is, \(1\) and \(G\) are the only subgroups of \(G\).

Next, we show that \(G\) is finite. Consider a non-identity element \(x \in G\). Since \(\langle x \rangle \ne 1\) and \(\langle x \rangle \le G\), by what we showed earlier, \(\langle x \rangle = G\).

Now consider \(\langle x^2 \rangle\). \(\langle x^2 \rangle\) is also a subgroup of \(G\), so it must equal either \(1\) or \(G\). If \(\langle x^2 \rangle = 1\), then \(|x| \le 2\) and thus \(G = \langle x \rangle\) is finite. If, on the other hand, \(G = \langle x \rangle = \langle x^2 \rangle\), then there must exist an integer \(k\) such that \(x = x^{2k}\). That again means that \(G = \langle x \rangle\) is finite.

Since \(G\) is finite and \(|G| > 1\), we can let \(|G| = pn\) for some prime \(p\) and positive integer \(n\). By Theorem 21, there exists \(g \in G\) where \(|g| = p\). Using the same argument we used above, \(G = \langle g \rangle\) since \(\langle g \rangle \le G\) while \(\langle g \rangle \ne 1\). Therefore, \(G \cong Z_p\).

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