Proving $N_2$ is not normal subgroup of $H_2$ if $\phi$ is not surjective

Question

I am given that $\phi: H_1 \to H_2$ is a non-surjective group homomorphism and $\phi(N_1) = N_2$ where $N_1 \unlhd H_1$. How do I prove that $N_2$ may not be a normal subgroup of $H_2$?

Attempt: Just thinking of an example which proves the statement. But I cannot come up with one. Can anyone give me a hint?

Answer 1

Take $H_1= \Bbb Z$ and $H_2=S_3.$ Let $\sigma \in S_3$ such that $\text {ord} (\sigma)=2.$ Consider a map $\varphi : H_1 \longrightarrow H_2$ defined by $$\varphi (i) = {\sigma}^{i},\ i \in \Bbb Z.$$ Observe that $\varphi$ is a non-surjective group homomorphism. What is $\varphi (\Bbb Z)$? Is it normal in $S_3$?

Answer 2

Take $H_2$ to be your favourite non-abelian simple group. If $H_2$ was chosen well then it contains a non-simple subgroup. Take this subgroup to be $H_1$, and $\phi$ to be the embedding map $\phi: H_1\hookrightarrow H_2$.

As a concrete example, $A_5$ contains the Klein $4$-group.

Answer 3

To add to Dbchatto67's answer: An overall idea of coming up with a counterexample is as follows:

1. Let $H_1$ be a cyclic group of non-prime odd order $q$ (and generator $\alpha$) so that it has a normal subgroup.

2. Let $H_2$ be a simple group with an element $\sigma$ of order $q$ i.e., the alternating group $A_q$.

Then $\phi: H_1 \mapsto H_2$ where $\phi(\alpha^i) = \sigma^i$ is well-defined, and furthermore, $\phi(H_1)$ is isomorphic to $H_1$; it is not a normal subgroup in the larger group $A_q$.

In general let $H_1$ be any group with a normal subgroup and let $H_2$ be any group that has as a subgroup $H'_1$ that satisfies (a) $H'_1$ isomorphic to $H_1$ and (b) $H'_1$ is not normal in $H_2$. Then the isomorphism from $H_1$ to $H'_1$ will work.