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Homomorphism of symmetry group

kalish

Member
Oct 7, 2013
99
I have a question that I have approached, but want to check if I'm on the right track.

Let G denote the group of symmetries of a circle. There are infinitely many reflections and rotations. There are no elements besides reflections and rotations. The identity element is the rotation by zero degrees. Let H denote the subgroup of H consisting of only the rotations.

Question: Define f: G--> {1,-1} by f(x) = 1 if x is a rotation and f(x) = -1 if x is a reflection. Is f a homomorphism?

What I did was check if f(xy)=f(x)f(y). So f((-1)*1)=f(-1)f(1)=-1, and f((1)*1)=f(1)f(1)=1, etc. It all checks out, so f is a homomorphism.

Did I do it right? Thanks!!
 

eddybob123

Active member
Aug 18, 2013
76
I have a question that I have approached, but want to check if I'm on the right track.

Let G denote the group of symmetries of a circle. There are infinitely many reflections and rotations. There are no elements besides reflections and rotations. The identity element is the rotation by zero degrees. Let H denote the subgroup of H consisting of only the rotations.

Question: Define f: G--> {1,-1} by f(x) = 1 if x is a rotation and f(x) = -1 if x is a reflection. Is f a homomorphism?

What I did was check if f(xy)=f(x)f(y). So f((-1)*1)=f(-1)f(1)=-1, and f((1)*1)=f(1)f(1)=1, etc. It all checks out, so f is a homomorphism.

Did I do it right? Thanks!!
A homomorphism between two groups is a function $f:G\to H$ that satisfies
$$f(g_1\cdot_Gg_2)=f(g_1)\cdot_Hf(g_2)$$
where $g_1,g_2\in G$, $\cdot_G$ is the group operation in G, and $\cdot_H$ is the group operation in H. The way you performed your calculations looks like you have mistaken the group operation in the symmetry group to be multiplication.
 

kalish

Member
Oct 7, 2013
99
So... is this the correct approach?

We need to show that f(x) f(y) = f(xy) for any x, y in G.

We prove this by cases:
(i) x and y are rotations. Then, xy is also a rotation.
So, f(x) f(y) = 1 * 1 = 1 = f(xy).

(ii) x and y are reflections. Then, xy is a rotation.
So, f(x) f(y) = -1 * -1 = 1 = f(xy).

(iii) x is a reflection and y is a rotation. Then, xy is also a reflection.
So, f(x) f(y) = -1 * 1 = -1 = f(xy).

(iv) x is a rotation and y is a reflection. Then, xy is also a reflection.
So, f(x) f(y) = -1 * 1 = -1 = f(xy).
 

eddybob123

Active member
Aug 18, 2013
76
I don't know what the best way to solve this problem is. What I would do is the following: first, we know that the sign homomorphism which maps each permutation to its sign, is indeed a homomorphism. Now try to prove that the symmetry group of a circle is isomorphic to a permutation group (which turns out to be of arbitrary large degree).
 

Evgeny.Makarov

Well-known member
MHB Math Scholar
Jan 30, 2012
2,492
What I would do is the following: first, we know that the sign homomorphism which maps each permutation to its sign, is indeed a homomorphism.
Isn't the sign of a permutation defined only for finite sets?

Now try to prove that the symmetry group of a circle is isomorphic to a permutation group (which turns out to be of arbitrary large degree).
There are definitely more permutations (i.e., bijections) of a circle than rotations and symmetries. Besides, this approach would require proving that the sign of a symmetry is -1 and the sign of a rotation is 1 or vice versa.

In my opinion, the solution in post #3 is fine.
 

eddybob123

Active member
Aug 18, 2013
76

Deveno

Well-known member
MHB Math Scholar
Feb 15, 2012
1,967
Isn't the sign of a permutation defined only for finite sets?

Now try to prove that the symmetry group of a circle is isomorphic to a permutation group (which turns out to be of arbitrary large degree).
There are definitely more permutations (i.e., bijections) of a circle than rotations and symmetries. Besides, this approach would require proving that the sign of a symmetry is -1 and the sign of a rotation is 1 or vice versa.

In my opinion, the solution in post #3 is fine.
I would also agree with this, PROVIDED:

you can PROVE that:

a) two rotations composed yield a rotation
b) a rotation and reflection composed in any order yield a reflection
c) two reflections composed yield a rotation.

In my humble opinion, one needs a specific "form" of rotations and reflections to do this. One possible approach:

DEFINE a rotation as a linear map:

$T(x,y) = (x\cos\alpha - y\sin\alpha ,x\sin\alpha + y\cos\alpha)$

for some real number $\alpha \in [0,2\pi)$

Similarly, DEFINE a reflection as:

$T \circ R$, where:

$T$ is a rotation, and; $R(x,y) = (x,-y)$

(other definitions are possible, be aware of this).

Note that to prove (a) through (c) using these definitions one must find some "angle" that works for the composition.

********

Also, in my humble opinion, the EASIEST way to prove the original assertion of post #1 is by noting this is just a special case of the homomorphism:

$\det:\text{GL}_2(\Bbb R) \to \Bbb R \setminus\{0\}$

restricted to the subgroup $\text{O}_2(\Bbb R)$ consisting of all linear plane isometries (which of necessity must map any circle centered at the origin (including the unit circle) to itself), using the multiplicative property of the determinant map.