Alternate approach using general dihedral group theory.
Recall standard facts about $D_n = \langle r,s \mid r^n=e, s^2=e, srs^{-1}=r^{-1}\rangle$, the symmetry group of a regular $n$-gon, of order $2n$:
$\bullet$ $D_n$ is non-abelian for every $n\ge 3$, since $srs^{-1}=r^{-1}\ne r$. With $n=5$, $D_5$ is non-commutative, so option (A) is true.
$\bullet$ $|D_n|=2n$ always, so $|D_5|=2(5)=10$, confirming option (B).
$\bullet$ The center of $D_n$ is $\{e\}$ when $n$ is odd, and $\{e,r^{n/2}\}$ when $n$ is even. Since $n=5$ is odd, $Z(D_5)=\{e\}$, confirming option (C).
$\bullet$ For normal subgroups: the rotation subgroup $\langle r\rangle$ (index 2) is always normal, and when $n$ is odd every reflection lies in a single conjugacy class of size $n$, so no reflection subgroup is normal. The complete list of normal subgroups of $D_5$ is therefore $\{e\}$, $\langle r\rangle$, and $D_5$ itself: exactly 3 normal subgroups, not 2.
So option (D)'s claim of 2 normal subgroups is incorrect, making it the statement that is NOT true.
\[\boxed{\text{(D)}}\]