# Help please. proof in algebra

1. Jul 14, 2011

### 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.

2. Jul 14, 2011

### micromass

Staff Emeritus
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

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

3. Jul 15, 2011

### moont14263

Thank you micromass, you are helpful.