Is $G/H$ isomorphic to $G'$ if there exists an homomorphism between $G$ and $G'$ and $H$ being a normal subgroup?

I have three questions :

I understand the first isomorphic theorem, which states that a homomorphic image of a group is isomorphic to the quotient group formed by the group $$G$$ and the kernel of group $$G$$.

$$1$$. Is this theorem only true for Kernel $$K$$, or for any normal subgroup of $$G$$ ?

Also, suppose there exists a homomorphism $$\phi$$ between $$G$$ and $$G'$$. Let $$H = \{x \in G \; ; \; \phi(x) \in H'\}.$$ Then $$H$$ is subgroup of $$G$$. We can also show that given that $$H'$$ is normal in $$G'$$, $$H$$ is normal in $$G$$. Here, there exists an homomorphism between $$H$$ and $$H'$$.

$$2$$. Is the function defining homomorphism between $$G$$ and $$G'$$ same as $$H$$ and $$H'$$ ?

From first isomorphism theorem, we can say, $$G/K \cong G'$$ and $$H/K \cong H'$$

$$3$$. Then Can I make this statement : Given a group $$G$$, and subgroup $$H$$ of $$G$$, if there exists a homomorphism between $$G$$ and $$G'$$ with Kernel $$K$$ and $$H'$$ being a subgroup of $$G'$$, such that $$G/K \cong G'$$ and $$H/K \cong H'$$, then $$H$$ is normal in $$G$$ and $$H'$$ is normal in $$G'$$ ?

To answer your first question, the kernel depends on the homomorphism in question, with a different homomorphism you'll have a different kernel. However, given any normal subgroup $$H$$, you can always find a homomorphism to some group whose kernel is $$H$$, namely, $$\phi\colon G\to G/H$$ is a surjective homomorphism with kernel $$H$$.

I don't quite understand your second question. I'm assuming you have a homomorphism $$\phi\colon G\to G'$$ and $$H'\lhd G'$$. If you are taking $$H = \phi^{-1}(H')$$, then $$H$$ is the kernel of $$G\xrightarrow{\phi} G'\to G'/H',$$ hence it is normal. You can restrict $$\phi$$ to $$H$$ to get a surjective homomorphism $$\phi\colon H\to H'$$ and from the first isomorphism theorem, $$H/K\cong H'$$.

Lastly, your claim is not true. Simply take a subgroup $$H$$ which is not normal, and the identity map $$G\to G$$. Then the kernel $$K$$ is trivial, $$G/K\cong G, H/K\cong H$$ but $$H$$ is not normal.

• Regarding first question, I understand that there will be a homomorphism between G and G/H. My question is when H is just a normal subgroup and not kernel of G, then will there be an isomorphism between G/H and G'(homomorphic image of G) ? Oct 21, 2020 at 5:42
• @latus_rectum No, not in general. For example, consider a finite group, then $G/H$ and $G'\cong G/K$ need not even have the same cardinality. Also, it is possible that $G/H\cong G/K$ via some isomorphism, but this isomorphism need not be related to the given $\phi\colon G\to G'$, i.e., there may be an isomorphism, but it need not arise in some natural way from $\phi$. Oct 21, 2020 at 6:51
• And clarity regarding second question :: There exists a homomorphism between G and G' given by $\phi$. And from mapping which I have mentioned in question, we have established that there exists a homomorphism between H(subgroup of G) and H'(subgroup of G'). My question is : Is this homomorphism also defined by using the same function $\phi$ or is it some other homomorphic function ? Oct 21, 2020 at 7:12
• @latus_rectum what is the mapping you mention? It should be the restriction of $\phi$ to $H$ as I have written in my answer and well, this is the only way we have to get from $H$ to $H'$. Oct 21, 2020 at 7:40
• @latus_rectum well, if you have subgroups $H, H'$ of $G, G'$ respectively, then there can be a homomorphism between them, independent of $G, G'$ or any homomorphism between $G, G'$. Oct 22, 2020 at 2:39