TITLE: How is this subgroup normal?
QUESTION [7 upvotes]: Let $G$ be a group, and let $U$ be a subset of $G$. Let $\hat{U}$ be the smallest subgroup of $G$ containing $U$. Then $\hat{U}$ is the intersection of the collection of all the subgroups of $G$ containing $U$. (Right?) This collection is obviously non-empty as $G$ itself is a subgroup of itself which contains $U$. 
Now my question is this: 
If $gug^{-1} \in U$ for all $g \in G$ and $u \in U$, then is the subgroup $\hat{U}$ a normal subgroup of $G$? 
If the answer is yes, then one way of proving this is to find a homomorphism $ \phi \colon G \to G$ with kernel equal to $\hat{U}$. 
If the answer is no, then what counter-example can we give?

REPLY [3 votes]: There's an explicit characterization of the smallest subgroup containing a given set:
$$\hat U=\{u_1^{\varepsilon_1}\ldots u_n^{\varepsilon_n}\mid u_i\in U,\varepsilon_i\in\{-1,1\}\text{ and }n\ge0\text{ (not a mistake)}\}$$
Because it must contain all these elements and it's closed under all group operations (including the identity).
Now observe that for any $g\in G$ we have
$$gu_1^{\varepsilon_1}\ldots u_n^{\varepsilon_n}g^{-1}=gu_1^{\varepsilon_1}g^{-1}\ldots gu_n^{\varepsilon_n}g^{-1}=(gu_1g^{-1})^{\varepsilon_1}\ldots(gu_ng^{-1})^{\varepsilon_n}={u'_1}^{\varepsilon_1}\ldots{u'_n}^{\varepsilon_n}$$
So $\hat U$ is a normal subgroup of $G$.