# Thread: Group of symmetries of a circle

1. ## Group of symmetries of a circle

Hello all,
I'm brand new to the forum and my name is Kalish. I'm from Russia. I have some questions on abstract algebra and differential geometry. Hope you guys will be able to help! Thanks in advance.

Problem statement: Let D_infinity denote the group of symmetries of a circle. Let SD_infinity denote the subgroup of D_infinity consisting of only the rotations. (By the way, SD_infinity is isomorphic to R/Z and to the complex unit circle.)

How can I find [D_infinity:SD_infinity]? (The index of SD_infinity in D_infinity). Is it 2, because the number of cosets is just the identity and the composition of rotations on the flips?

Also, how can I prove that D_infinity is not isomorphic to SD_infinity x {-1,1}? What group-theoretic property do they NOT share, such that I can separate the two? I'm stumped on this one.

2. ## Re: Group of symmetries of a circle

It seems to me that you can choose any axis to flip the circle. The axis you choose would determine the points that remain fixed by the flip. If this is true, then for any flip, multiplying by the rotation $\displaystyle e^{i\pi}$ would give the identity. So, the flips could be uniquely identified by an element of $\displaystyle \{e^{i\theta}\mid \theta \in [0,\pi)\}$ (so half the points of the circle, since every flip uses two points).

3. ## Re: Group of symmetries of a circle

So are you saying that the flips and rotations are basically the same?

4. ## Re: Group of symmetries of a circle

I think I was wrong. There is no rotation so that a flip and a rotation gives the identity. Consider a flip across the x-axis. Hence $\displaystyle e^{i\theta} \mapsto e^{-i\theta}$. There is no rotation that will return all points to their original position. But, for any flip there is a rotation so that the rotation followed by a flip across the x-axis, then the inverse rotation is the same as just the original flip.

5. ## Re: Group of symmetries of a circle

So, it would make sense to say that there is only one flip (since any other flip is the same as a rotation, a flip, then an inverse rotation). So, we have the one flip is $\displaystyle s: e^{i\theta} \mapsto e^{-i\theta}$. Then the rotation $\displaystyle r_\psi: e^{i\theta} \mapsto e^{i(\theta+\psi)}$. Then the word $\displaystyle r_\psi s r_\psi^{-1} s^{-1} = \mbox{Id}$.

6. ## Re: Group of symmetries of a circle

So, yes, it seems clear that the index of $\displaystyle [D_\infty : SD_\infty] = 2$ since $\displaystyle s^2 = \mbox{Id}$. To show that $\displaystyle D_\infty$ is not isomorphic to $\displaystyle SD_\infty \times \{-1,1\}$, how does multiplication work in the latter group? Then assume $\displaystyle \varphi: D_\infty \to SD_\infty \times \{-1,1\}$ is an isomorphism. Try taking a word through the morphism and show that something fails? Off the top of my head, I am guessing something relating to the relationship $\displaystyle sr = r^{-1}s$ that might fail.

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