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Let $\varphi:G\to G'$ be a group homomorphism. Is $\varphi(G)$ a normal subgroup of $G'$?

I dont know how to prove that the statement is false.

Without lose of generality suppose that $G$ is a subgroup of $G'$ and $\varphi$ is the inclusion map, then the problem is reduced to prove that not any subgroup is normal.

Then, how I can prove that not any subgroup is normal? I dont know any example or how I can derive this result from the axioms of groups. Can you help me please? Thank you.

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    $\begingroup$ You could(should) replace the question with "Is $\phi(G)$ necessarily a normal subgroup of $G'$." $\endgroup$ – Corey Sep 18 '16 at 10:34
  • $\begingroup$ You should also try to come up with rules for addition that would make a group have noncommutative elements. Then you could start coming up with counter examples on your own. $\endgroup$ – Corey Sep 18 '16 at 10:36
  • $\begingroup$ @Corey I changed the title to something that I think is the real question. $\endgroup$ – Masacroso Sep 18 '16 at 10:37
  • $\begingroup$ The yellowed portion of the post was the problem from your book? That doesn't seem to fit with the title of your post.. $\endgroup$ – Corey Sep 18 '16 at 10:38
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    $\begingroup$ It is uncommon in mathematics to reserve an adjective for a property that always holds; you can therefore safely assume that if a notion of "normal" subgroup is defined, not all subgroups are normal. You could therefore have asked "please show me an example of a subgroup that is not a normal subgroup". But if your book/course defines normal subgroups, they ought to give examples both of subgroups that are normal and of subgroups that are not, in which case you could have spared asking the question by looking that up. $\endgroup$ – Marc van Leeuwen Sep 18 '16 at 12:34
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This is wrong in general. For instance, let $G=\langle (1 \; 2)\rangle$, which is a non-normal subgroup of $G'=S_3$ (let $g=(1 \; 3) \in G'$, and show that $g^{-1}(1 \; 2)g \not \in G$), and just take $\phi : G \to G'$ the inclusion.

However, if $H$ is a normal subgroup of $G$, and if $\phi : G \to G'$ is a surjective group homomorphism, then $\phi(H)$ is a normal subgroup of $G'$.

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For (counter) example:

$$G=\{\,(1),\,(12)\,\}\;,\;\;G'=S_3\;,\;\;\phi:G\to G'\;,\;\;\phi(12)=(12)\;,\;\;\phi(12)=(1)$$

In fact, you can choose any group $\;G'\;$ which has at least one non-normal subgroup $\;G\;$ , and then take the inclusion homomorphism $\;G\to G'\;$ and you get a counter example.

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  • $\begingroup$ What mean $S_3$? Sorry, Im starting with group theory. $\endgroup$ – Masacroso Sep 18 '16 at 10:17
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    $\begingroup$ The group of permutation on three letters $\endgroup$ – DonAntonio Sep 18 '16 at 10:18
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Here is an infinite example. You can check that $2\times2$ matrices of non-zero determinant with complex entries form a group for matrix multiplication. Denote this by $G'$. Now consider the complex numbers of modulus 1 which is a group for multiplication of numbers. Denote this by $G$.

Now define $\phi\colon G\to G'$ by $\displaystyle \phi(z) = {z\quad 0\choose 0\quad \bar z}$. You can check that $\phi(G)$ is not a normal subgroup. [Use the matrix ${1\quad1\choose 0\quad1}$.]

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It is known the alternating group $A_n$ is simple, i.e. has no normal subgroup except the trivial ones, for all $n\ge 5$.

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    $\begingroup$ Maybe not the best answer for somebody who has apparently never seen a non-normal subgroup. $\endgroup$ – Marc van Leeuwen Sep 18 '16 at 12:37
  • $\begingroup$ Maybe. I admit it has the unique advantage of brevity. $\endgroup$ – Bernard Sep 18 '16 at 12:44
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    $\begingroup$ This is one of the most confusing answers you could have given. $\endgroup$ – Lily Chung Sep 18 '16 at 15:33

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