## Subgroups of dihedral groups (2)

Posted: February 17, 2011 in Elementary Algebra; Problems & Solutions, Groups and Fields
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For the first part see here. I will follow the following notation:

Notation. $D_{2n} =\{a^rb^s : \ 0 \leq r < 2, \ 0 \leq s < n \},$ with $o(a)=2, \ o(b)=n$ and $ab=b^{-1}a.$

Lemma 3. Given $d \mid n,$ let $m = n/d.$ For every $0 \leq i < n$ let $A(i,d) = \{ab^{i+km}: \ 0 \leq k < d \}.$ Let $B(i,d) = A(i,d) \cup \langle b^m \rangle.$ Then $B(i,d)$ is a subgroup of $D_{2n}$ and $|B(i,d)|=2d.$ We also have $|\{B(i,d) : \ 0 \leq i < n \}|=m.$

Proof. If $ab^{i+km}=ab^{i+rm},$ for some $0 \leq k,r < d,$ then $b^{(k-r)m}=1$ and thus $d \mid k-r,$ because $o(b)=n=md.$ Therefore $k=r$ because $0 \leq k,r < d.$ So $|A(i,d)|=d.$ Clearly $A(i,d) \cap \langle b^m \rangle = \emptyset$ and $|\langle b^m \rangle| = d,$ because $o(b)=n.$ Thus $|B(i,d)|=|A(i,d)| + |\langle b^m \rangle|=2d.$ Proving that $B(i,d)$ is a subgroup of $D_{2n}$ is easy. Just note that every element of $A(i,d)$ is the inverse of itself (because they all have order two) and also note that $ab^s = b^{-s}a,$ for all $s,$ because $ab=b^{-1}a.$ Finally, the set $\{B(i,d) : \ 0 \leq i < n \}$ has $m$ elements because clearly $B(i,d)=B(j,d)$ if and only if $A(i,d)=A(j,d)$ if and only if $i \equiv j \mod m. \ \Box$

Theorem. (Stephan A. Cavior, 1975) If $n \geq 3,$ then the number of subgroups of $D_{2n}$ is $\tau(n) + \sigma(n).$

Proof.  Suppose that $H$ is a subgroup of $D_{2n}.$ There are two cases to consider.

Case 1 . $H \subseteq \langle b \rangle.$ By Lemma 1, the number of these subgroups is $\tau(n).$

Case 2 . $H \nsubseteq \langle b \rangle.$ In this case, by  Lemma 2, we have $|H|=2d$ and $|H \cap \langle b \rangle|=d,$ for some $d \mid n.$ Let $n = md.$ Since $H \cap \langle b \rangle$ is a subgroup of $\langle b \rangle,$ which is a cyclic group of order $n,$ we have

$H \cap \langle b \rangle = \langle b^m \rangle. \ \ \ \ \ \ \ \ \ \ (*)$

Let $A(i,d)$ and $B(i,d)$ be as they were defined in Lemma 3. Now, since $H$ is not contained in $\langle b \rangle,$ there exists some $0 \leq i < n$ such that $ab^i \in H.$ Then, since $H$ is a subgroup, we must have $ab^ib^{km} \in H,$ for all $k.$ Thus $ab^{i + km} \in H$ and so $A(i,d) \subseteq H$ and therefore, by $(*),$ we have $B(i,d) \subseteq H.$ Thus, since $|H|=|B(i,d)|=2d,$ we must have $H=B(i,d).$ The converse is obvously true, i.e. given $d \mid n$ and $0 \leq i < n,$ $B(i,d)$ is a subgroup of $D_{2n},$ by Lemma 3, and $B(i,d) \nsubseteq \langle b \rangle$ because it contains $A(i,d).$ So the subgroups in this case are exactly the ones in the form $B(i,d),$ where $0 \leq i < n$ and $d \mid n.$ Thus, by Lemma 3, the number of subgroups in this case is

$\sum_{d \mid n} | \{B(i,d) : \ 0 \leq i < n \} | = \sum_{d \mid n} n/d = \sum_{d \mid n} d = \sigma(n).$

So, by case 1 and case 2, the number of subgroups of $D_{2n}$ is $\tau(n) + \sigma(n). \ \Box$

Note that we didn’t just find the number of subgroups of $D_{2n}.$ We also found all the subgroups.

Example. Find all subgroups of $D_{12}.$

Solution. There are $\tau(6)+\sigma(6)=4+12=16$ subgroups. Four of them are obtained from case 1 in the proof of the theorem. They are the subgroups of $\langle b \rangle.$ Since $o(b)=6,$ the subgroups in this case are $\{1\}, \ \langle b \rangle, \ \langle b^2 \rangle$ and $\langle b^3 \rangle.$ There are $12$ subgroups left and they are in the form $B(i,d),$ where $0 \leq i < 6$ and $d \mid 6.$ So $d = 1, 2,3$ or $6.$ Also, by the proof of the last part of Lemma 3, $B(i,d)=B(j,d)$ iff $i \equiv j \mod 6/d.$ So those $12$ subgroups are:

$B(0,1), B(1,1),B(2,1),B(3,1),B(4,1),B(5,1), B(0,2),B(1,2),B(2,2),B(0,3),B(1,3),B(0,6).$

Note that $B(0,6)=D_{12}.$