Normal subgroup generated by a singleton Let $G$ be a group and $a\in G$. Let $H$ be the normal subgroup of $G$ generated by $a$: that is,
$$H=\bigcup_{n\in\mathbb{N}}\left\{y\ \Biggm|\ (\exists g)(\exists x)(g\in G\land x\in\{a,a^{-1}\}^{[1,n]}\land y=\prod_{i=1}^ngx_ig^{-1})\right\}.$$
Is there a simpler description of $H$ since we are generating with only one element?
Edit:
Set $X=\{y\in G\ |\ (\exists g)(g\in G\land y=gag^{-1})\}$. Then the desired subgroup is 
$$\bigcup_{n\in\mathbb{N}}\left\{y\ \Biggm|\ (\exists x)(x\in (X\cup X^{-1})^{[1,n]}\land y=\prod_{i=1}^ngx_ig^{-1})\right\}?$$
 A: First, both given descriptions are wrong, even setting aside the appearance of unquantified variables and the disappearance of quantified ones.
Your first attempt makes $H$ a union of products where each factor is either $gag^{-1}$ or $ga^{-1}g^{-1}$, for a fixed $g$. Since $(gxg^{-1})(gyg^{-1}) = g(xy)g^{-1}$, these products just amount to $ga^kg^{-1}$ for some integer $k$. Your $n$ also ensures that $-n\leq k\leq n$, but then taking the union means that your set $H$ (not a subgroup!) is the union of the conjugacy classes of powers of $a$. While all of these elements lie in $\langle a\rangle^G$, the normal closure of $\langle a\rangle$, they are not that subgroup: the set may not be closed under products. For example, if $g\neq h$, there is no reason to expect the product $(gag^{-1})(hah^{-1})$ to lie in your set. 
Your second attempt is slightly better, but still incorrect. Now you are trying to take products of conjugates of $a$ and their inverses. Note that now your $g$ is unquantified and unnecessary. I suspect you meant to take products of conjugates of $a$ and their inverses. If so, then that would have been accurate, but your description is lacking. Perhaps you would need to do the following: let $X$ be
$$X = \{gag^{—1}\mid g\in G\}$$
the conjugacy class of $X$, and then let
$$H = \bigcup_{n\in\mathbb{N}}\left\{y\Biggm|\Bigl(\exists x_1,\ldots,x_n\in X\cup X^{-1}\Bigr) \Bigl( y = \prod_{i=1}^n x_i\Big)\right\}$$
that is, all finite products of conjugates of $a$ and their inverses. That description is correct.
Now, to answer your question, that’s pretty much the best you can do. It describes it as $\langle a^G\rangle$, the subgroup generated by the conjugacy class of $a$. The only alternative is to describe it as the product of conjugates of elements of $\langle a\rangle$, that is, expand your $X$ to include elements of the form $ga^ig^{-1}$ with $i\in\mathbb{Z}$, in which case the description for $H$ can become
$$H = \bigcup_{n\in\mathbb{N}}\left\{y\Biggm|\Bigl(\exists x_1,\ldots,x_n\in X\Bigr) \Bigl( y = \prod_{i=1}^n x_i\Big)\right\}$$
omitting $X^{-1}$ from the pool from which you take the elements, since $(ga^kg^{-1})^{-1} = ga^{-k}g^{-1}$. 
If there is a simpler formula-description, I am not aware. Wordwise, “products of conjugates of powers of $a$” does the trick. 
