Question

If $f:G\to H$ is a surjective homomorphism of groups and if N is a normal subgroup of G, prove that $f\left(N\right)$ is a normal subgroup of H .

Short answer

It is proved that f(N) is a normal subgroup of H .

Step-by-step solution

Given in the question

We know from the question that:

• $f:G\to H$ is surjective homomorphism
• N is a normal subgroup of G

Proving that f(N)  is a normal subgroup of  H.

Since $f:G\to H$ is a surjective homomorphism, therefore, for any arbitrary $a\in H$, and $b\in G$ we can write: $f\left(b\right)=a$

Suppose, f(N) is a normal subgroup of H.

Then, for $n\in N$: $af\left(n\right)a^{-1}\in f\left(N\right)$

$f\left(b\right)f\left(n\right)f{\left(b\right)}^{-1}\in f\left(N\right)$

Since we know $f:G\to H$ is a surjective homomorphism

$f\left(b\right)f\left(n\right)f{\left(b\right)}^{-1}$=$f\left(bn{b}^{-1}\right)$

Therefore,

$f\left(bn{b}^{-1}\right)\in f\left(N\right)$

This implies that $bn{b}^{-1}\in N$. This is true because N is a normal subgroup in G.

This proves that our assumption is true. Hence, f(N) is a normal subgroup of H .