## Computation of the number of Sylow 2-subgroups in Dih(2n)

Let $2n = 2^ak$ where $k$ is odd. Prove that the number of Sylow 2-subgroups of $D_{2n}$ is $k$.

We begin with several lemmas.

Lemma 1: If $n = km$, then $D_{2m} \leq D_{2n}$. Proof: The homomorphism defined on the generators of $D_{2m}$ by $s \mapsto s$ and $r \mapsto r^k$ is injective. $\square$

Lemma 2: Let $G$ be a finite group and $p$ a prime. If $H \leq G$ is a subgroup such that $p$ does not divide $[G:H]$, then every Sylow $p$-subgroup of $H$ is a Sylow $p$-subgroup of $G$. Proof: We have $|G| = p^am$ and $|H| = p^bn$, with $p$ not dividing $m$ and $n$ and $n|m$; since $p$ does not divide the index of $H$ in $G$, $a=b$. Hence any Sylow $p$ subgroup of $H$ has order $p^a$, and thus is a Sylow $p$-subgroup in $G$. $\square$

Now to the main result.

Suppose $n$ is odd. Then $k = n$ and every Sylow 2-subgroup has order 2. In this case, the Sylow 2-subgroups correspond to the elements of order 2 in $D_{2n}$. There are exactly $n$ of these: elements of the form $sr^b$ for $0 \leq b < n$. Thus $n_2 = k$.

Suppose now that $n = 2 \ell$ is even; then $a \geq 2$. We proceed by induction on the width of $k$. (Recall that the width of an integer is the number of prime factors which divide it, including multiplicity.) Let $P$ be a Sylow 2-subgroup of $D_{2n}$.

For the base case, if the width of $k$ is 0, we have $k = 1$. Then $P = D_{2n}$. In this case, $P$ is the unique Sylow 2-subgroup, so that $n_2 = 1$.

For the inductive step, suppose that for any integer $k$ of width at most $t$, the number of Sylow 2-subgroups of $D_{2^ak}$ is $k$. Let $k$ be an integer of width $t+1$. Then $k = qm$ for some prime $q$ and some width $t$ integer $m$; without loss of generality, we may assume that $q$ is the smallest prime divisor of $n$, so that $q \leq m$. By the induction hypothesis, $D_{2^am}$ has $m$ distinct Sylow 2-subgroups. By Lemma 1, $D_{2^am} \leq D_{2^aqm}$. Note also that 2 does not divide $[D_{2^aqm} : D_{2^am}] = q$, so that by Lemma 2, $D_{2^am}$ has at least $m$ distinct Sylow 2-subgroups.

Now by Sylow’s Theorem, $n_2$ divides $k = qm$. Thus either $n_2 = m$ or $n_2 = k$. Note that there exist elements of order 2 in $D_{2n}$ which are not in (the image of) $D_{2m}$; for instance, $sr$, since otherwise we have $(sr^m)(sr)(r^m) = r$ in the image of $D_{2m}$ in $D_{2n}$, a contradiction. By Sylow’s Theorem, every 2-subgroup (For instance $\langle sr \rangle$) is contained in some Sylow 2-subgroup. Thus there must exist at least one other Sylow 2-subgroup in $D_{2n}$; hence $n_p = k$.

By induction, the result is proved for all $D_{2n}$.