If the set of regular languages is closed under the concatenation operation and is also closed under the reverse operation ($x^R$ is the reverse of $x$) then is the language generated by $$\{ww^R|w\in\Sigma^*\}$$ for some input alphabet $\Sigma$, also regular? If not, why not?
I've been trying to find a proof for this using the pumping lemma, but it seems that selecting any substring towards the middle of the string being pumped could also be of the form $\{ww^R|w\in\Sigma^*\}$, causing the original string to remain in its original form.
Here's a try:
$\textbf{Theorem:}$ The language, $A$, generated by $\{ww^R|w\in\Sigma^*\}$ is not regular.
$\textbf{Proof:}$ Assume $A$ is regular (We will use the Pumping Lemma for Regular Languages to show a contradiction). Let the input string $s$ be $ww^R$ and let $p = |w|$.
When splitting $s$ into substrings $x, y, z$ such that $s=xyz$ we see that $xy$ must be a substring of $w$ by the third condition of the Pumping Lemma ($|xy|\le p$).
By the first condition of the Pumping Lemma, we see that all strings of the form $xy^iz$ must be in $A$ for all $i \ge 0$. Taking $i$ to be zero, we obtain the string $xw^R$. $|x| < |w^R|$ so $xy^0z \notin A$.
QED? What if $xw^R$ can still be split so that for some substring $k$, $kk^R = xw^R$?
I think I may be overthinking this but it's really bugging me.