Why is $\mathbb{S}^1$ not a normal subgroup of $\mathbb{S}^3$

Question

I know in general we try to show the left cosets are not equal to the right cosets but I need something more concrete to get me started in this example. How can I begin? (Here, $\mathbb{S}^3$ is the 3-sphere), can be written as 2x2 complex matrices, i.e. group of unit quaternions.

Answer 1

There are several equivalent formulations of normality. You might find some other formulation more useful than the words "left cosets are right cosets".

To start with, you can re-express those words in actual mathematics: for a subgroup $A$ of a group $G$ to be normal means

1. For all $x \in G$ we have $xA=Ax$.

You can then also rewrite this as

2. For all $x \in G$ we have $xAx^{-1}=A$.

So, to prove this is false, find a counter-example: take $G = \mathbb{S}^3$, take $A = \mathbb{S}^1$, and start trying different values of $x$.

Answer 2

I assume that $\mathbb{S}^3$ is the group of unit quaternions, while $\mathbb{S}^1$ is the group of unit complex numbers. The quaternion $i$ can be conjugated to any purely imaginary (that is, of the form $a i + b j + c k$) unit quaternion, as you can glean from this Wikipedia article: http://www.wikiwand.com/en/Quaternions_and_spatial_rotation

Answer 3

So in other words:

$$S^1:=\{a+bi\in\Bbb H\;;\;\;a^2+b^2=1\}\;,\;S^3:=\{a+bi+cj+dk\in\Bbb H\;;\;\;a^2+b^2+c^2+d^2=1\}$$

and $\;\Bbb H\;$ is the algebra of quaternions.

But check that

$$\underbrace{\left(\frac1{\sqrt2}(1-j)\right)^{-1}}_{\frac1{\sqrt2}(1+j)}\cdot i\cdot\left(\frac1{\sqrt2}(1-j)\right)\notin S^1$$