Proving that the generator of $U$ is normal if $\forall u \in U, g\in G$ $gug^{-1} \in U$ This is from Herstein. 

$4.$ $\;a)$ Given a group $G$ and a subset $U$ denote by $\hat U$ the
  smallest subgroup of $G$ which contains $U$ (the subgroup generated by
  $U$). Prove there is a subgroup $\hat U$ of $G$.  
$\;\;\;\;\;b)$ If $gug^{-1} \in U$ for all $g \in G$ and $u \in U$,
  prove that $\hat U$ is a normal subgroup of $G$.

My Work: Let $\mathscr A$ be the collection of subgroups of $G$ that contain $U$. This collection is non-empty since it includes $G$ itself. I defined the subgroup generated by $U$ to be, $$ \hat U = \bigcap_{A \in \mathscr A} A $$
Then we can prove that $\hat U$ is a subgroup of $G$, it contains $U$ and if $H$ is any subgroup of $G$ that contains $U$ then $\hat U$ is a subgroup of $H$. 
But I am having trouble proving part $b)$. Any help is appreciated. A hint would be fabulous. 
Thanks in advance.  
 A: Hint:  Show that
$$V:=\{v\in \hat{U}: gvg^{-1}\in \hat{U},\forall g\in G\}$$
is a subgroup containing $U$. It is clear that $V\subseteq\hat{U}$. Also by minimality of $\hat{U}$, it follows that $V\supseteq\hat{U}$. Hence $V=\hat{U}$, this means for all $v\in\hat{U}$ we have $gvg^{-1}\in\hat{U},\forall g\in G$, which means $\hat{U}$ is normal.
A: First show that for every $g\in G$ and $A\in \mathscr A$, $gAg^{=1}\in \mathscr A$.
Indeed,
$$A\in \mathscr A\Rightarrow U\subset A\Rightarrow U=gUg^{-1}\subset gAg^{-1}.$$
Now
$$\hat U=\bigcap_{A\in\mathscr A}A\subset gAg^{-1}$$
for every $A\in \mathscr A$. Hence
$$\hat U\subset\bigcap_{A\in\mathscr A} gAg^{-1}=g\hat Ug^{-1}.$$
This can also be written as
$$g^{-1}\hat Ug\subset \hat U.$$
Replacing $g$ by $g^{-1}$ gives
$$g\hat Ug^{-1}\subset\hat U.$$
Since we have inclusions bothways, the identity is proved.
A: First we prove that $\forall \hat u \in \hat U$, $\exists u, v \in U$, s.t. $\hat u = uv$.
To do so we define set $\hat W = \{ uv | uv \in \hat U, u, v \in U \}$, and prove $\hat W$ is exactly $\hat U$.
Consider $\forall w_1, w_2 \in \hat W$.
By definition of $\hat W$, we have $w_1, w_2 \in \hat U$. Since $\hat U$ is a group, $w_1 w_2^{-1} \in \hat U$.
By definition of $\hat W$, we write $ w_1 = u_1 v_1$, $w_2 = u_2 v_2$. So $w_1 w_2^{-1} = (u_1 v_1) (u_2 v_2)^{-1} = (u_1 u_2) (v_2^{-1} v_1^{-1})$. 
Since $U$ and $\hat U$ are group, $u_1 u_2, v_2^{-1} v_1^{-1} \in U$. So by definition of $\hat W$, $w_1 w_2^{-1} \in \hat W$.
So $\hat W$ is also a group. 
But $U \subset \hat W$, and $\hat U$ is the smallest group that contains $U$, we have $\hat W = \hat U$.
With this we can proceed to

b) If $gug^{−1} \in U$ for all $g\in G$ and $u\in U$, prove that $\hat U$ is a normal subgroup
  of G.

We just proved above that $\forall \hat u \in \hat U$, there $\exists u, v \in U$ s.t. $\hat u = uv$.
So $g\hat u g^{-1} = g uv g^{-1} = (g u g^{-1}) (g v g^{-1})$, where $gug^{-1}, gvg^{-1} \in U$, so $g \hat u g^{-1} \in \hat U$.
So $\hat U$ is normal.
