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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?

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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 $S_3$?

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  • $\begingroup$ I know this might be a weird question, but what is the thought-process when it comes to question like these? This is my first time studying group theory, and I am struggling with it. It is just amazing that you can cook up an example in less than 3 minutes. Does this come with practice? $\endgroup$ – Ufomammut Apr 18 at 17:06
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    $\begingroup$ Some kind of experience is definitely needed. But what I did can come to anyone's mind. Because to get a non-normal subgroup of $H_2$ we must need the non-abelianness of $H_2.$ The most common non-abelian group which came into our mind is obviously $S_3.$ Now we know that $S_3$ has only one proper normal subgroup of order $3.$ So to cook up a non-normal subgroup you have to definitely go for the cyclic subgroups generated by the transpositions. Right? That is essentially what I did. Nothing amazing in it. $\endgroup$ – Dbchatto67 Apr 18 at 17:14
  • $\begingroup$ Is the above statement true in general? $\endgroup$ – Ufomammut Apr 18 at 17:18
  • $\begingroup$ Which statement are you trying to say? $\endgroup$ – Dbchatto67 Apr 18 at 17:31
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    $\begingroup$ Nope. It can't be. $\Bbb Z$ contains odd as well as even integers. Can $\varphi$ map to any odd integer? $\endgroup$ – Dbchatto67 Apr 18 at 17:57
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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.

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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.

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