I am going through Algebra In Action by Shahriari and on page 147 it says
Definition 7.8 (Sylow subgroups). Let G be a finite group with |G| = $p^a$m, where
$p \nmid m$, and $a$ is a non-negative integer. A subgroup $P \le G$ with $|P| = p^a$ is called
a Sylow $p$-subgroup of $G$.
The set of all Sylow $p$-subgroups of $G$ is denoted by $\operatorname{Syl}_p(G)$.
Example 7.9. If $G$ is a group of order $24$, then a subgroup of order $3$ would be
one of its Sylow $3$-subgroups and a subgroup of order $8$ would be one of its Sylow
$2$-subgroups. The identity subgroup would be its Sylow $5$-subgroup as well as its
Sylow $11$-subgroup. While we know that $G$ has the identity subgroup, we need
Sylow’s theorem to know that $G$ actually must have a subgroup of order $3$ and a
subgroup of order $8$.
Why is the identity subgroup a $5$-subgroup and an $11$-subgroup??!!
EDIT $5^0 = \{$identity of $G\}$ is considered a sylow 5 group by this definition.
Best Answer
$5$ and $11$ do not divide $24$, i.e. their highest powers which divide $24$ are $5^0$ and $11^0$. So the corresponding Sylow sub-groups have $5^0 = 11^0 = 1$ element. What's a subgroup with just one element? The identity subgroup!
Note that in the definition, $\alpha \ge 0$ and not just $\alpha > 0$. So, $\alpha = 0$ is perfectly okay to consider.