Let $A$ be an alphabet.
By $A^{**}$ let us denote the set of all strings from $A^*$ of even length. (This definition may be incorrect but it was given to me in the question)
Show that $$(AA)^{*} = A^{**}$$
My Answer:
I believe I need to show two cases, not sure if I proved the cases exactly correct.
- $(AA)^{*} \subseteq A^{**}$
- $A^{**} \subseteq (AA)^{*}$
case 1: Suppose $w \in (AA)^{*}$ then by definition of Kleene star $w$ will contain $0$ or more strings of $AA$ which means $w$ can be a string from $A^*$ since $A^*$ just repeats the string with a longer length and concatenation with two strings from $AA$ is the same as a string repeated with longer length so
$$w \in A^{**} \implies (AA)^{*} \subseteq A^{**}$$
case 2: Suppose $w \in A^{**}$ then $w$ is a $0$ or more strings from the alphabet $A$ and adding another Kleene star would make more combinations of the strings from $A$ with higher length which is the same as concatenating two strings to equal the same length so
$$w \in (AA)^{*} \implies A^{**} \subseteq (AA)^{*}$$
Much of your trouble with this problem seems to come from the poor choice of notation by the author: $A^{**}$ is not a good way to denote the set of even-length strings. Pre-empting the result we are asked to prove, one might suggest $A^{2*}$.
The key to understanding the theorem, as well as where you went wrong in your proposed solution, is to understand what a string of elements of $AA$ looks like.
The way you put it:
suggests that you interpret $(AA)^*$ as being $A^*A^*$, i.e. the concatenations of two strings in $A$.
However, let us explicitly expand the definition of Kleene star:
$$(AA)^* = \{w_1 w_2 \ldots w_n: n \ge 0, \forall n: w_n \in AA\}$$
Now ask yourself the following questions: