Describe Sylow subgroups of $D_4$ I should describe Sylow subgroups of $D_4$.
$|D_4| = 8 = 2^3$, so $D_4$ has just 2-subgroup.
But I don't understand, how can i prove it formally.
 A: Let's go through the various parts of Sylow's theorems, one by one, and try to understand how they apply to $D_4$.

Part 1: If $|G| = p^n m$, where $p$ is prime and $p$ does not divide $m$, then $G$ has a subgroup of order $p^n$. (Any such subgroup of order $p^n$ is called a Sylow $p$-subgroup of $G$.)

As you correctly pointed out, $ D_4$ has order $2^3$, so $p = 2$ is the only prime that divides the order of $D_4$. And $2^3$ is the highest power of $2$ that divides the order of $D_4$. Therefore, a Sylow $2$-subgroup of $D_4$ is a subgroup of order $2^3$. Well, does $D_4$ contain a Sylow $2$-subgroup? Yes it does: the entire group $D_4$ is (trivially) a subgroup of $D_4$ of order $2^3$. So Part 1 is satisfied.

Part 2: All Sylow $p$-subgroups of $G$ are conjugate subgroups.

Well, the entire group $D_4$ is actually the only Sylow $2$-subgroup of $D_4$. And any subgroup is trivially conjugate to itself.

Part 3: Suppose $|G| = p^n m$, and let $n_p$ be the number of distinct Sylow $p$-subgroups of $G$. Then $n_p$ divides $m$ and $n_p \equiv 1 {\rm \ mod \ } p$.

In our example, we have $p = 2$, $n = 3$, $m = 1$ and $n_2 = 1$. And yes, $ 1$ divides $1$, and $ 1$ is congruent to $1$ modulo $2$. So everything works!
