Let A be a regular set . Consider the two sets below.

L1 = {$x \mid \exists{n}\geq 0 , \exists{y} \in A : y =x^{n}$}

L1 = {$x \mid \exists{n}\geq 0 , \exists{y} \in A : x =y^{n}$}

Which of the following is True ?

 1. L1 and L2 are Regular
 2. L1 is Regular but Not L2
 3. L2 is Regular but Not L1
 4. Both are not Regular

**My Friend explained to me like this** 

Let us consider the language L2 first.Given that A is a regular set and the string denoted by 'y' $\in$ A , so we know that power which we take as concatenation of the string itself  "n times " .

We are taking power of a language as concatenation "n" times with itself because :

Given a set V define

V0 = {ε} (the language consisting only of the empty string),

V1 = V

and define recursively the set

Vi+1 = { wv : w ∈ Vi and v ∈ V } for each i>0.

So Vi+1 is nothing but set comprising of concatenation of w and v where w belongs to Vi which I am referring as Vi  and v belongs to V. 

Reference : Definition and Notation Part of https://en.wikipedia.org/wiki/Kleene_star

Now the set $V$ in this question is referred to the regular set A. We know that concatenation of regular language (or) regular set results in regular language only(by closure properties of regular language) .

So X is generated by Y's concatenation only and $Y$ is a regular set (or) language.So $X$ is also going to be regular set.

Now coming to language L1.

Now it says the opposite i.e. $Y \in A $ only but now the relation between Y and X is :  $Y  =  X^{n}$  and given the clause there exist associated with the value of $n$ , so we can assign any value of n which is $>= 0$ .So if $n = 1$ , then $Y = X$ and hence $X$ is obviously regular set .

Similarly on setting $n = 2$ , we get $Y = X^{2}$  meaning Y can be partitioned on exactly 2 halves and hence we can say X is half(Y).And we know given a language or a set L is regular , then half(L) is also regular.

**Answer given as L1 is regular and NOT L2**

What is the correct answer ? How to solve this kind of questions ?