Let f: G -> K be an epimorphism of groups. If N is a normal
subgroup of G, show that f(N) := {f(n) | n e N} is a normal subgroup of K.

please note in this proof "e" is the id for G
and "i" is the id for K
My proof:
Let g, h be an elements of G
Suppose N is a normal subgroup of G,

then f(e)=i is element of f(N)
f(g^(-1))= f(g)^(-1) is element of f(N)
f(g)f(h) = f(gh)=f(N)
therefore N is subgroup of K

Now to proof that it is a normal subgroup of K

f(ghg(-1))=f(g)f(h)f(g)^(-1)

Now im stuck...
Can some one please show me how to proof this , and if my proof that i gave so far
is incorrect plz show me how in the simplest way ...

Thank you in advance.

2. Originally Posted by Dreamer78692
Let f: G -> K be an epimorphism of groups. If N is a normal
subgroup of G, show that f(N) := {f(n) | n e N} is a normal subgroup of K.

please note in this proof "e" is the id for G
and "i" is the id for K
My proof:
Let g, h be an elements of G
Suppose N is a normal subgroup of G,

then f(e)=i is element of f(N)
f(g^(-1))= f(g)^(-1) is element of f(N)
f(g)f(h) = f(gh)=f(N)
therefore N is subgroup of K

Now to proof that it is a normal subgroup of K

f(ghg(-1))=f(g)f(h)f(g)^(-1)

Now im stuck...
Can some one please show me how to proof this , and if my proof that i gave so far
is incorrect plz show me how in the simplest way ...

Thank you in advance.

$\displaystyle \forall k_1\in K\,\exists g_1\in G\,\,s.t.\,\,f(g_1)=k_1$ , so that if $\displaystyle k\in f(N)$ then also $\displaystyle k=f(g)$ for some $\displaystyle g/in G$, and now:

For any elements $\displaystyle k_1\in K\,,\,k\in f(N)$ , $\displaystyle k_1^{-1}kk_1=f(g_1)^{-1}f(g)f(g_1)=f(g_1^{-1}gg_1)$ , and since $\displaystyle g_1^{-1}gg_1\in N$ then $\displaystyle k_1^{-1}kk_1\in f(N)$ and we're done.

Tonio