A dihedral group is a group of symmetries of a regular polygon, with respect to function composition on its symmetrical rotations and reflections, and identity is the trivial rotation where the symmetry is unchanged. For such an \(n\)-sided polygon, the corresponding dihedral group, known as \(D_{n}\) has order \(2n\), and has \(n\) rotations and \(n\) reflections.
Let us denote the first rotation by \(y\). Rotating the symmetry \(n\) times would again give the same position of the symmetry. So performing function composition \(y\), \(n\) times, we get the identity \(e\), therefore we can write it as \(y^{n}\ = e\). Also, let us denote the first reflection by \(x\). Reflecting the symmetry twice will give us symmetry with the original position. Thus, we denote \(x^{2} = e\), and similarly, every reflection has order 2.
Dihedral groups are generated by a reflection and a rotation, such that we write dihedral group \(D_{n}=< y, x >\) (finitely generated). Here \(y^{n} = e\) and \(x^{2} = e\). So, we get rotations of the form of integer powers of \(y\). For reflections, we have them in the form of composition of \(x\) with some power of \(y\), i.e., \(x{y}^{k}\), \(k\) is between 0 and \(n − 1\) (both inclusive). Thus, \[D_{n} = {e, y, y^{2}, . . . , y^{n−1}, x, xy, x{y}^{2}, . . . , x{y}^{n−1}}, and |D_{n}| = 2n,\]
where \(y^{n} = e\) and \(x^{2} = e\).
We can notice that the rotations in \(G = D_{n}\) form a group in itself.
So we take \(H = {e, y, y^{2}, y^{3}, . . . , y^{n−1}}\) where \(y^{n} = e\).
So \(H < G\)(\(H\) is the subgroup of \(G = D_{n}\)). Clearly, \(H =< y >\) is a cyclic group since all the elements
of \(H\), i.e., the rotations, are some integral powers of \(y\). Thus, \(|H| = n\).
If we find the index of \(H\) in \(G\), then it would be \(\frac{2n}{n} = 2\). Since \(H\) comes out to be a subgroup of Index-2, then we can say that any arbitrary subgroup of \(G\) will either be contained in \(H\) or will have exactly half of the elements as that in \(H\). If \(S < G\) is any arbitrary subgroup contained in \(H\), then \(S < H\), and since \(H\) is cyclic, \(S\) will also be cyclic.
Otherwise, if \(S < G\) isn’t contained in \(H\), then it has exactly half rotations. Since \(S < G\), it can have only rotations and reflections. Since there are half rotations in \(S\), the other half are reflections in \(S\). Thus, \(S\) is a dihedral group.
Any subgroup of a dihedral group is either cyclic group or dihedral group.
For any arbitrary \(S\ <\ G\), if \(S\) is contained in \(H\), then \(S\ <\ H\), and \(S\) is cyclic. Since \(H\ =<\ y\ >\) and subgroup \(S\ <<\ y\ >\), thus, \(S\) will be generated by some integer power of \(y\) where the integer power divides \(n\), because \(H\ =<\ y\ >\) and \(y^{n} = e\), and thus, \(S\ =<\ y^{n/d}\ >\) where \(d|n\) (\(d\) divides \(n\)), for \(1 \leq d \leq n\). For such \(y^{n/d}\), clearly the order is \(d\). Thus, \(y\) raised to the power of divisors of \(n\) generate \(S\), for all possible divisors. So, there are as many as cyclic subgroups as there are divisors of \(n\).
For any arbitrary \(S < G\), if \(S\) is not contained in \(H\), then \(S\) is a dihedral group. So, \(S\) will be generated
by a rotation and a reflection. For a reflection, we initially have \(n\) choices: \(x, xy, x{y}^{2}, . . . , x{y}^{n−1}\). For
the rotation, since a rotation will again generate a cyclic subgroup within the new dihedral subgroup,
and this new cyclic subgroup will certainly be a subgroup of \(H\), and as \(H\ =<\ y\ >\), therefore, the
rotation can be of the form \(y^{n/d}\) only, where \(d|n\). Here, as mentioned above, the order of \(y^{n/d}\) is \(d\),
and thus there are \(d\) rotations in \(S\), and being a dihedral the same number of reflections also. So, we
conclude that:
\(S\ =<\ y^{n/d},\ x{y}^{p}\ >\) for \(d|n\) and \(0 \leq p < n.\)
For a given \(d|n\), when we try generating \(S\) with the resulting \(y^{n/d}\) and with all possible values of \(p\) between 0 and \(n − 1\), taken one by one, we find that we obtain exactly \(\frac{n}{d}\)distinct dihedral groups (implying that there are exactly \(\frac{n}{d}\) choices for \(p\)). This is because the dihedrals obtained when we take \(x{y}^{p}\) and \(x{y}^{p+i(n/d)}\), where \(i\) is some non-negative integer, are the same.
The dihedrals obtained when we take \(x{y}^{p}\) and \(x{y}^{p+i(n/d)}\) are same because for some \(p\), we obtain a dihedral by performing the operation \(x{y}^{p} · {y}^{u(n/d)}\) for non-negative integer \(u\), thus getting \(x{y}^{p+u(n/d)}\). The same dihedral is obtained when we have \(x{y}^{p+i(n/d)}\), and perform operation \(x{y}^{p+i(n/d)} · {y}^{t(n/d)}\) for some non-negative integer \(t\), thus making the resultant as \(x{y}^{p+(i+t)(n/d)}\) where \(i + t\) is again a non-negative integer, which is equivalent to \(x{y}^{p+u(n/d)}\). So basically, for each \(d|n\), we have exactly \(\frac{n}{d}\) number of dihedral subgroups, which is equivalent to saying that for \(d\) number of rotations and \(d\) number of reflections, we have exactly \(\frac{n}{d}\) distinct such dihedral groups in \(D_{n}\). So, for every divisor of \(n\), there are that number of dihedrals present, which means that the total number of dihedral subgroups is the sum of all positive divisors of \(n\).
Number of dihedral subgroups of \(D_{n}\) = sum of all positive divisors of \(n\) in \(D_{n}\)
As \(1|n\), so for \(d\ =\ 1\) in the above setup, we get \(S\ =\ <\ {y}^{n},\ x{y}^{p}\ >\), or \(S\ =<\ x{y}^{p}\ >\), making \(S\) cyclic. As there are \(\frac{n}{1}\ =\ n\) choices for \(p\), we conclude that there are \(n\) number of cyclic dihedral subgroups. So interestingly, they show both natures: dihedral because they have 1 rotation and 1 reflection, and cyclic because their order is 2.
Number of subgroups of \(D_{n}\) which are both dihedral and cyclic in nature \(= n\)
Now, we aim to find the exact number of subgroups present in \({D}_{n}\) for any given order. For any arbitrary \(S\ <\ G\), let us have \(|S|\ =\ s\). If \(S\ <\ H\), then \(s|n\). Since there exists a unique subgroup of each possible order in a cyclic group, therefore, we have 1 cyclic subgroup for order \(s\) of \(S\) where \(s|n\). If \(s\) does not divide \(n\), then we don’t have this subgroup. On the other hand, if \(S\) isn’t contained in \(H\), then \(S\) is dihedral of order \(s\) with \(\frac{s}{2}\) rotations and \(\frac{s}{2}\) reflections in \(S\). If \(s\) is odd, \(\frac{s}{2}\) is not an integer and therefore \(S\) does not exist. So for \(s\) being odd, we have no dihedral subgroups in \(D_{n}\).
If \(s\) is even, then it is possible to have \(\frac{s}{2}\) rotations and reflections. From previous argument, we had that for \(d\) number of rotations and \(d\) number of reflections, there were exactly \(\frac{n}{d}\) distinct dihedral subgroups, and thus in analogy with this result, we get \(\frac{n}{s/2}\) distinct dihedral subgroups, i.e., \(\frac{2n}{s}\) dihedral subgroups.
So if \(s|n\), we have a cyclic subgroup. If \(s\) doesn’t divide \(n\), then no such cyclic subgroup is present. Also, if \(n\) is even, we have \(\frac{2n}{s}\) dihedral subgroups, but for \(n\) being odd, we don’t have any dihedral subgroup in \(D_{n}\).
For any \(S\ <\ G\) where \(|S|\ =\ s\), we have following four cases:
Example: In case of \(D_{6}\), possible orders are 1, 2, 3, 4, 6, 12. Total there are 4 cyclic and 12 dihedral subgroups.
For \(s = 1\), there is only 1 subgroup (The trivial Identity group).
For \(s = 2\), there are 7 subgroups.
For \(s = 3\), there is only 1 subgroup.
For \(s = 4\), there are 3 subgroups.
For \(s = 6\), there are 3 subgroups.
For \(s = 12\), there is only 1 subgroup (The Group itself).
Designed & Developed by: Vedant Goyal and Ujjawal Agarwal