Sylow p-subgroup is p-group Let P be a prime. A p-group is a group of order $P^{r} \forall r\geq 0$.
If G is a group of order $P^{r}m \forall m \in \mathbb{Z}$ where p does no divide m then a Sylow p-subgroup of G is a subgroup of order $p^{r}$.

From this, would readers be valid in making the deduction that a Sylow p-subgroup is a p-group?
Thanks in advance.
The deduction would be valid, since you seems to allow for the possibility of $r=0$ for both p-groups and Sylow-p-subgroups. 

However, my textbook (Artin) defines p-groups such that $r\geq1$. If we defined Sylow-p-subgroups such that trivial Sylow-p-subgroups would exist (like my professor does), then not all Sylow-p-subgroups would be p-subgroups if we combine Artin's definition with my professor's definition.
The deduction would be valid, since you seems to allow for the possibility of $r=0$ for both p-groups and Sylow-p-subgroups. 

However, my textbook (Artin) defines p-groups such that $r\geq1$. If we defined Sylow-p-subgroups such that trivial Sylow-p-subgroups would exist (like my professor does), then not all Sylow-p-subgroups would be p-subgroups if we combine Artin's definition with my professor's definition.