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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 $G'$ ?

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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.

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  • $\begingroup$ 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) ? $\endgroup$ Oct 21, 2020 at 5:42
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    $\begingroup$ @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 $\phi$. $\endgroup$ Oct 21, 2020 at 6:51
  • $\begingroup$ 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 ? $\endgroup$ Oct 21, 2020 at 7:12
  • $\begingroup$ @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 $H'$. $\endgroup$ Oct 21, 2020 at 7:40
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    $\begingroup$ @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 $G, G'$. $\endgroup$ Oct 22, 2020 at 2:39

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