Given a group $G$ and two subgroups $N,K$ and its inclusion map : $$j_{1} : N \to G$$ $$j_{2} : K \to G$$ Extend to the homomorphism : $$j : N*K \to G$$ Let $A$ be a normal subgroup of $N*K$ and $A \subset Ker j$ and the map induced by $j$ : $$ k : (N*K)/A \to G$$ Is an isomorphism then $A = Ker j$ .
Normal subgroup equal kernel
0
$\begingroup$
group-theory
-
0You mean the statement $A = Ker j$ is true ? – 2017-01-24
-
0I mean $k$ is an isomorphism and I need $A = Ker j$ – 2017-01-24
-
0OK, sorry I misunderstood. If $g \in {\rm Ker}(j)$ then $gA \in {\rm Ker}(k)$. So if $k$ is an isomorphism then $gA = A$, hence $g \in A$, and so $A = {\rm Ker}(k)$. – 2017-01-24