subgroups that contain a normal subgroup is also normal Let $F$ be a field and $G$ the set of degree 1 polynomials in $F[x]$. We showed that $G'$ the set of elements of the form $x + p$ where $p \in F$ is a normal subgroup. Now suppose, $S \subset G$ is a normal subgroup, if $S \neq \{e\}$ show that $G' \subset S$. And also, on the converse, if $S$ is a subgroup of $G$, and $S$ contains $G'$ show that $S$ is a normal subgroup of $G$. 
I tried defining $S = \{cx + d : c \neq 1 \}$ and computing $gSg^{-1}$, find a contradiction, but I couldn't. Any direction would be appreciated! 
 A: Consider $f:G\rightarrow F$ defined by $f(ax+b)=a$, it is a morphism of groups and its kernel is $G'$, s $G'$ is normal.
Let $S$ be a normal subgroup and $ax+b\in S$  distinct of the identity,  suppose that there exists $ax+b=x+p\in S$, $p$ distinct of $0$, 
let $q\in F$ ${q\over p}x\circ (x+p)\circ {p\over q}x={q\over p}x\circ ({p\over q}x+p)=x+q$ implies that $G'\subset S$.
Suppose that $a\neq 1$.
$ ax\circ (ax+b)\circ{1\over a}x=ax\circ(x+b)=ax+ba\in S$, $(ax+b)^{-1}\circ (ax+ab)=({1\over a}x-{b\over a})\circ (ax+ab)=x+b-{b\over a}$, $b-{b\over a}=0$ implies that $b(a-1)=0$ since $a\neq 1$, we deduce that $b=0$.
Let $p\neq 0$ $(x+p)\circ ax\circ (x-p)=(x+p)\circ (ax-ap)=ax-ap+p\in S$, we deduce that $(ax-ap+p)\circ(ax)^{-1}=x-ap+p\in S$ which is not trivial since $a\neq 1$, as above, we deduce that $G'$ is a subgroup of $S$.
Suppose that $S$ is a subgroup of $G$ which contains $G'$, $f(S)$ is a normal subgroup of $F-\{0\}$ since $F-\{0\}$ is commutative, it implies that $S=f^{-1}(f(S))$ is a normal subgroup of $G$.
