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

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?

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 $$S_3$$?

• 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? – Ufomammut Apr 18 at 17:06
• 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. – Dbchatto67 Apr 18 at 17:14
• Is the above statement true in general? – Ufomammut Apr 18 at 17:18
• Which statement are you trying to say? – Dbchatto67 Apr 18 at 17:31
• Nope. It can't be. $\Bbb Z$ contains odd as well as even integers. Can $\varphi$ map to any odd integer? – Dbchatto67 Apr 18 at 17:57

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.

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.