(February 2013) Let \(G_1\) and \(G_2\) be groups, let \(\varphi:G_1\to G_2\) be a homomorphism, and let \(H_1\subseteq G_1\) be a subgroup. Recall that the set \[ H_2=\{\varphi(x)\mid x\in H_1\} \] is a subgroup of \(G_2\), called the image of \(H_1\) under \(\varphi\), sometimes notated \(\varphi(H_1)\). If \(G_1\) is finite, prove that \(|H_2|\mid |G_1|\). That is, prove that the order of \(H_2\) divides the order of \(G_1\).
Solution
Proof. Let K = Ker(φ) denote the kernel of φ. Then: • φ(G1)∼=G1/K by the fundamental theorem of group homomorphisms, so |φ(G1)| =|G1/K|. • We also know that|G1/K| =|G1|/|K|. Together, these two facts yield (1) |φ(G1)|·| K| =|G1|, and hence|φ(G1)| ⏐⏐⏐|G1| We also observe the following about H1 and H2: • Since H1⊆G1, we have H2 =φ(H1)⊆φ(G1). • By Lagrange’s Theorem, it follows that |H2| divides|φ(G1)|. Since|H2| divides|φ(G1)| by the last bullet, and |φ(G1)| divides|G1| by (1), it follows that |H2| divides|G1|. QED