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Let ∑ = {a} and L1 = ∑* 

Then we can say that L1 is regular.

Now take any arbitrary string x belongs to ∑*

then x = am for some integer m

Now, m must have a prime factor p

Hence m = p*k (for some integer k)

Hence, am = ap . ap . ap . ap .....(concatenating the string ap k times)

Now, ap belongs to L (given in the question )  which means (ap )k  belongs to L*

Hence am belongs to L*  which means x belongs to L*

So, for any arbitrary string x belongs to ∑* we get that x belongs to L*

Hence ∑* = L* 

Thus L* is a regular language.

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Subham Nagar asked May 4, 2018
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I don't think it will be regular .Had the language been uwwRv the expression could have been (a+b)*(aa+bb)(a+b)* as there is no restriction on w.Is it correct? and if it ...