Question

Every homomorphic image of a finite group is finite

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ItsSamGupta

Let ϕ:G→G′ be a homomorphism. Show that if |G| is finite, then |ϕ[G]| is finite and divides |G|.

Because φ[G]={φ(g)|g∈G}, we see |φ[G]|≤|G| which the question presupposed is finite. By Theorem 13.15, there is a one-to-one correspondence between the elements of φ[G] and the cosets of Ker(φ) in G.

(1.) How do you envisage and envision to use Theorem 13.15? How does it relate to this question?

Thus |φ[G]|=|G|/|Ker(φ)|, so |φ[G]| divides |G|.

(2.) Where does |φ[G]|=|G|/|Ker(φ)| crop up from? No isomorphism theorems are covered.

(3.) What's the intuition? Another proof

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Jan 26 '14 at 8:36

Nai

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Apr 13 '17 at 12:19

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φ partitions G into n equivalence classes, one for each preimage of an element of φ(G). Do we know that each equivalence class is the same size, m? Then certainly the equivalence class that is the preimage of the only element we unquestionably know is in G′, the identity, (i.e., kerφ) has size m. Then since G is partitioned by these equivalence classes, |G|=n⋅m=|φ(G)| |kerφ|.