# Finding cosets and index in D8

a). Find all the left and right cosets of $<τ>$ in $D_8$.

$<τ>$ is reflection and I know that there are two reflections about a diagonal. So I am wondering how to represent this idea as left and right cosets. My book did not go into detail about cosets or give any examples.

b). Find the index of $<ρ^2 τ>$ in $D_8$.

So from my understanding, $D_8$ is a group of order $8$ which is representative of a $4-gon$. So then there are $19$ total symmetries. There should be a total of two reflections about a single diagonal axis with rotation by $\pi$ from the first diagonal. My question for this is how do I find the index from this? My book did a poor job of defining index and cosets. I know I have to find the number of left cosets of $<ρ^2 τ>$ in $D_8$ but I don't know how to show this.

Any help would be appreciated.

"D8" is not "a 4-gon". D8 is a group with eight elements, one possible interpretation of which is the group of rotations and reflections of a square. There are not "19 elements" in the group, there are, as I said, 8. There are not "two reflections about an axis", there are two axes with one reflection about each. There are also two more reflections, about the lines through the center parallel to two of the sides.

Now, what are $\tau$ and $\rho$?

• $τ$ would be a reflection about axes $2$ and $4$ of the shape and $ρ$ is a rotation by angle $\pi$.Thank you for clearing that up. So I understand that I have a group with order 8. When I said 19 elements, I think I was trying to say that I have 19 symmetries of a 4-gon. So where do I go from here? – TfwBear Oct 27 '15 at 1:01

I'd recommend rereading your text, since you seem to be confused by what cosets are. I don't know how bad your text is, but if it's that bad, you can always examine Wikipedia's summary to get an idea of what's happening.

For a brief overview, given a group G and a subgroup of G, H, a left coset of H in G is a set of the form {a * h: h ∈ H, a ∈ G}. Similarly, a right coset of G in H is a set of the form {h * a: h ∈ H，g ∈ G}.

Since we're working with a relatively small group, you can just check the Cayley table yourself for the results. For an approach, start with e: you'll get the coset {e*h: h ∈ }, which is of course just . I don't know how far you've gotten in your text, but hopefully you'll know that left cosets form an equivalence relation among the group elements, so elements in the same left coset will generate the same left coset (though being in the same left coset does not generate being in the same right coset).

What that means for your answer is that once you've found a coset, you know the coset formed by all elements of that coset, and so you don't need to do the multiplications for each element, only the ones you have yet to put in a coset. A simple, brute force algorithm would be:

1. Let S be a list of group elements of G.
2. Pick an arbitrary element of S (though e is the most obvious)
3. Find and record the left coset given by that element for H in G.
4. Remove all elements of that coset from S.
5. Repeat steps 1 through 4 until S is empty.

The index is the number of left (or right) cosets given by a subgroup. Since each coset has cardinality equal to the size of the subgroup, if G is finite, the index of H is just |G|/|H|, so for your case, you can just find |D8|/|<ρ2r>|.