# Mathematics Prelims

## January 26, 2009

### Generators and the Dihedral Groups

Filed under: Abstract Algebra,Algebra — cjohnson @ 10:34 pm

Consider the Dihedral group of order 2n, denoted $D_{2n}$.  This group represents the “symmetries” of a regular n-sided polygon in the plane, where a “symmetry” is a way of moving the n-gon around with rigid body transformations in 3-space, and laying it back down perfectly on top of a copy of the original n-gon.  (Note that we’re using $D_{2n}$ to mean the symmetries of an n-gon, not a 2n-gon, consistent with Dummit & Foote.  Other authors, however, will use the notation $D_n$ for this group.  The reason for the 2n will become apparent soon.)

For instance, imagine an equilateral triangle in the plane.  Call one vertex A, another B, and third C.  We want to pick this triangle up and move it around so that when we lay it back down it completely covers the space occupied by the original triangle.  In doing so, though, the position of our A, B, C vertices may change.  For instance, if we picked the triangle up and placed it back down so that it had been rotated clockwise 120 degrees, instead of seeing the vertices A, B, C in their original order, we may now see C, A, B in their place.  We could also “pinch” one of the vertices, keeping it still, and rotating  the triangle so that the other two vetices swapped.  In this case we may go from A, B, C to A, C, B if A was the vertex we kept still.

Of course, we could also rotate the triangle some multiple of 120 degrees clockwise or counter-clockwise, or we could choose to “pinch” another vertex when we flip the triangle over as discussed above.  For instance, if we pinched B we might go from A, B, C to C, B, A.  Supposing we always pinch the top-most vertex, though, we could rotate, flip, then rotate back and achieve the same effect.  Any counter-clockwise rotation could likewise be represented as repeating the clockwise rotation an appropriate number of times.  For instance, if we rotate counter-clockwise we’d go from A, B, C to B, C, A.  If we just rotated clockwise twice though we’d go from A, B, C to C, A, B on the first rotation, then from there to B, C, A.  In general, any of these symmetries could be represented entirely as a combination of flips with the top-most vertex pinched, and clockwise rotations by 120 degrees.

What we’ve described for a triangle could easily be extended for polygons with more sides, though.  Suppose that we label the n vertices of our n-gon as 1, 2, 3, …, n clockwise around the figure.  Any of the symmetries we could perform is uniquely determined by two things:  which vertex we send 1 to, and whether 2 is “in front of” or “behind” 1 (meaning, if we were to traverse the figure’s vertices clockwise, if we’d get to 2 before we got to 1, or after).  This means that there are 2n possible symmetries:  pick a place to send 1 (n possibilities), then pick whether 2 comes before or after (2 possibilities).  This is the reason we use $D_{2n}$ instead of $D_n$ for this set.

Now, we’ve said that every element of $D_{2n}$ could be thought of as a combination of flips and rotations.  In algebra-speak, we say that $D_{2n}$ is generated by flips and rotations.  If we denote a flip by $f$ and a rotation by $r$, we would then write $D_{2n} = \langle f, r \rangle$ for this.  We can make this notion of a group being generated by a subset of its elements more precise, but before we do that we’ll need the following:

Let $G$ be a group, $\Lambda$ some index set, and suppose $H_\alpha \leq G$ for every $\alpha \in \Lambda$.  We will show that $\bigcap_{\alpha \in \Lambda} H_\alpha \leq G$ by applying the subgroup test previously described.  First note that the identity is in $\bigcap H_\alpha$ since each $H_\alpha \leq G$ means the identity is in each $H_\alpha$.  Now let $x,y \in \bigcap H_\alpha$.  Since $y \in \bigcap H_\alpha$, we must have $y \in H_\alpha$ for each $\alpha \in \Lambda$.  Since each of these is a subgroup we must have $y^{-1} \in H_\alpha$ for each $\alpha$, so $y^{-1} \in \bigcap H_\alpha$.  Now note $xy^{-1} \in H_\alpha$ for each $\alpha$, so $xy^{-1} \in \bigcap_{\alpha \in \Lambda} H_\alpha$, and $\bigcap H_\alpha \leq G$.  In short: an intersection of subgroups of some larger group is itself a subgroup of that group.

Now suppose $S$ is a general, non-empty subset of elements of $G$; $S$ need not be a subgroup of $G$.  The smallest subgroup containing $S$ (where “smallest” means with respect to the inclusion relation), which we call the subgroup generated by $S$ and denote $\langle S \rangle$, is the intersection of all subgroups of $G$ which contain $S$:

$\displaystyle \langle S \rangle = \bigcap_{\substack{S \subseteq H \\ H \leq G}} H$

We’ve shown that arbitrary intersections of subgroups forms another subgroup, and by construction $S$ is included in this subgroup.  If we have another subgroup containing $S$, though, it will be one of the $H$ subgroups in the above intersection, and so it will include this subgroup.  That is, if $S \subseteq H$, then $\langle S \rangle \leq H$.

Alternatively, we could describe the subgroup as the collection of all products, powers, and inverses of elements of $S$.  That is,

$\displaystyle \langle S \rangle = \{ t_1^{a_1} \cdot t_2^{a_2} \cdot ... \cdot t_n^{a_n} : n \in \mathbb{N}, \, t_i \in S, \, a_i \in \mathbb{Z} \, \text{ for each } 1 \leq i \leq n \}$

This tends to be an easier way of thinking about subgroups generated by a set: just mix up all the elements of the set however you want, and the collection of everything you could possibly get is the subgroup generated by the set.

In the case $S = \{s_1, s_2, ..., s_n\}$ we will normally write $\langle s_1, s_2, ..., s_n \rangle$ instead of $\langle \{ s_1, s_2, ..., s_n \} \rangle$, and if $S = \{s\}$ is a singleton set, we’ll just write $\langle s \rangle$.  A group generated by a single element is called a cyclic group, but that’s a topic for another post.