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Prove that the number of subgroups in $D_n = \tau (n) + \sigma (n)$ where $\tau (n)$ represents number of divisors of $n$ and $\sigma (n)$ represnts the sum of divisors of $n$.

Attempt: $D_n = \{e,r,r^2, \cdot \cdot \cdot, r^{n-1},s,rs,r^2s, \cdot \cdot \cdot, r^{n-1}s \}$

Then, $\{e,r,r^2, \cdot \cdot \cdot, r^{n-1}\}$ is a cyclic subgroup of $G$. There are $\tau(n)$ such cyclic subgroups.

Total number of subgroups of order $2=n+1$ if $n$ is even and $=n$ if $n$ is odd.

Hence, the sum is $\tau(n)+n$ or $\tau(n)+n+1$

Now, let $H$ denote the set of reflections, then if $a=r^is,b=r^js ~\in~H$, then $r^is~ (r^js)^{-1} = (r^is)^{-1}r^js=sr^{-i}r^js=sr^{j-i}s=s^2r^{i-j}=r^{i-j} \notin H$

Hence, the set of reflections cannot form a subgroup in themselves in any case.

Now, we must consider sets involving both rotations and reflections and find conditions for them to form subgroups.

By Lagrange's theorem, order of such a subgroup $H$ must divide $n$ as we have already found out the total number of subgroups of order $2$ So :

(a) A reflection is it's self inverse

(b) A reflection times a reflection is always a rotation

(c) Inverse of a rotation $r^i = r^{n-i}$

(d) A rotation times a rotation is always a rotation.

How do I choose such rotation and reflective elements such that they form a group. Suppose, $|H|=4$, In that case, $H$ should look something like this:

$H = \{f_1,f_2,r^i,r^{n-i}\}$ where $f_1,f_2$ are reflections and $f_1f_2=r^i$

How can I extend this to $n$ variables? A hint to move ahead would be really appreciated. Thanks

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Useful note prepared by Keith Conrad are [here](www.math.uconn.edu/~kconrad/blurbs/grouptheory/dihedral.pdf‎) and [here](www.math.uconn.edu/~kconrad/blurbs/grouptheory/dihedral2.pdf‎). –  user93432 May 3 at 9:23
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2 Answers 2

Hint: A dihedral group $D_n$ will contain other dihedral group $D_m$ as a subgroup if $m$ divides $n$.

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I am using your notation.

First, you have the subgroups of the cyclic subgroup $\langle r \rangle$ of order $n$. You should have seen that there are $\tau(n)$ such subgroups.

Now we are looking, as you did, at a subgroup $H$ that contains reflections. Suppose $H \cap \langle r \rangle = L$ has order $m \mid n$. Then it is easy to see that $H = \langle L, q \rangle$, where $q$ is any reflection in $H$. (This is because for such an $H$ you have $H \langle r \rangle = D_{n}$, so $D_{n}/ \langle r \rangle = H \langle r \rangle / \langle r \rangle \cong H / H \cap \langle r \rangle$ has order $2$.)

Given $L$, any reflection $q$ will determine such an $H = \langle L, q \rangle$, and any of the $m$ reflections of $H \setminus L$ determine the same $H$. Thus for each $L$ (that is, for each divisor $m$ of $n$), there are $n/m$ such subgroups $H$. It follows that the total number of such $H$ is $$ \sum_{m \mid n} \frac{n}{m} = \sigma(n). $$

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