Understanding Proof in Algebra: How to Prove H is Contained in N | Group Theory

[moont14263]

Can someone help to understand the under lined part of the proof:
Let G be a group and let N be a normal subgroup of G of finite index. Suppose that H is a finite subgroup of G and that the order of H is relatively prime to the index of N in G. Prove that H is contained in N

Proof:
Let f : G -> G/N be the natural projection. Then f(H) is a subgroup of G/N, so its order must be a divisor of |G/N|. On the other hand, |f(H)| must be a divisor of |H|. Since gcd (|H|,[G:N]) = 1, we must have |f(H)| = 1, which implies that H c ker (f ) = N.
Thanks in advance

 

[micromass]

This follows from the isomorphism theorem: for every group G and every homomorphism f, we have an isomorphism

$$f(G)\cong G/Ker(f)$$

thus

$$|G|=|f(G)||Ker(f)|$$

 

[moont14263]

Thank you micromass, you are helpful.