Homomorphisms and Normal Subgroups

Question

I have many questions regarding the normal subgroups and homomorphisms. Please help me to understand the concept.

$1$. Let $f:G \to G^{'}$ be a homomorphism between groups $G$ and $G'$.

By fundamental theorem of homomorphism the factor group $G/Ker(f)$ is isomorphic to $f(G)$. We know that Kernels are normal subgroup of $G$. So, for every homomorphism we get a kernel which is a normal subgroup in $G$. Is that right$?$ To be precise, for every homomorphism, can we get a normal subgroup in $G$?

2.If the above statement is true...then what about its converse..? (i.e) for every normal subgroup of the Group, can we construct a homomorphism? (irrespective of the codomain)

3.Let $K \subset H \subset G$ where $G$ is the group and $H$ and $K$ are subgroups of $G$...IS it true that if $K$ is normal in $G$ implies that $K$ is normal in $H$..? if it is false then under what condition it can be true..?

Answer 1

1. Yes. The kernel is always a normal subgroup of $G$.
2. The converse is also true. Given a normal subgroup $N$ of $G$, the map $G \rightarrow G/N$ has kernel $N$. Note that since $N$ is a normal subgroup, $G/N$ is a group.
3. $K$ is normal in $H$. It should be straightforward to verify this from definition.

Answer 2

2.If the above statement is true...then what about its converse..? (i.e) for every normal subgroup of the Group, can we construct a homomorphism? (irrespective of the codomain)

Well, the projection map $G \to G/N$ sending each element to its coset shows that every normal subgroup is indeed the kernel of a homomorphism. I'm not sure what you mean by irrespective of the codomain, but the codomain cannot be arbitrary. For example, if the codomain is the trivial group, then there is only one homomorphism from $G$, for which everything is in the kernel.

3.Let $K⊂H⊂G$ where $G$ is the group and $H$ and $K$ are subgroups of $G$...IS it true that if $K$ is normal in $G$ implies that $K$ is normal in $H$..? if it is false then under what condition it can be true..?

Yes this is true just by the definition of a normal subgroup. $K$ is normal in $G$ if $gkg^{-1} \in K$ for each $k \in K,g \in G$. but every element of $H$ is in $G$, so then $hkh^{-1} \in K$ for each $k \in K,h \in H$, so $K$ is normal in $H$.

Answer 3

For every homomorphism we get a kernel which is a normal subgroup in $G$.

$\textbf{2:}$ Let $N\unlhd G$. We construct natural homomorphism $\phi: G\rightarrow G/N$ such that $x\to xN$ and

ker $\phi=N$. Actually, this is important structure in maths.

• We can do it with equivalance relation in set theroy.
• We can do it with ideal in ring theory.
• We can do it with congruence in semigroup theory.
• Main idea is to define quotient what is kernel. according to me, the purpose here is to shred big structures.

$\textbf{3:}$ Yes. Look question $7$ here