Is any subgroup normal? 
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.
 A: 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 $G'$.
A: 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.
A: 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 ${1\quad1\choose 0\quad1}$.]
A: It is known  the alternating group $A_n$ is simple, i.e. has no normal subgroup except the trivial ones, for all $n\ge 5$.
