# Symmetry group

## Homework Statement

Show that symmetry operations for en greek vase build up a symmetry group.

## Homework Equations

For en greek vase we have
$$\Gamma=[e, C_{2},\sigma, \sigma^{'}]$$
And there are 3 conditions which must be fullfilled so that the elements will create a symmetry group $$1) (a\cdot b)\cdot c= a\cdot (b\cdot c)$$
$$2) a\cdot e= a$$
$$3) a\cdot a^{-1}=e$$

## The Attempt at a Solution

So we know that the vase is invariant under $$180^{0}$$ so it is of $$C_{2} type$$
do I understand correctly$$C_{2}\cdot C^{-1}_{2}=e$$ rotation $$180^{0}$$ and another one $$180^{0}$$ in the opposite direction
second condition-($$a\cdot e=a$$)
can we write then $$C_{2}\cdot e=C_{2}$$???

How will it work for the condition 1?$$(a\cdot b)\cdot c=a(b\cdot c)$$
Can we show it in this way? $$(C_{2}\cdot e)\cdot C^{-1}_{2}=C_{2}\cdot (e\cdot C^{-1}_{2})\rightarrow e=e$$
How can we show it with using other symmetry elements? $$[\sigma, \sigma{'}, e]$$

for example$$(C_{2}\cdot \sigma)\cdot e=C_{2}\cdot(\sigma\cdot e)$$???

## The Attempt at a Solution

fzero
Homework Helper
Gold Member
There's actually a condition that you must verify before considering the ones that you've listed. It is closure, that the product of two elements of $$\Gamma$$ is also an element belonging to $$\Gamma$$. Therefore, the first thing you want to do is build the multiplication table for the elements of $$\Gamma$$.

When you know the rules for multiplication, verifying associativity will be fairly simple.

you mean I have to calculate
$$C_{2}\cdot e=$$
$$C_{2}\cdot C_{2}=$$
$$C_{2}\cdot \sigma=$$
$$C_{2}\cdot \sigma^{'}=$$
$$\sigma^{'}\cdot \sigma{'}=$$
$$e\cdot e=$$
$$e\cdot C_{2}=$$
$$e\cdot \sigma=$$
$$e\cdot \sigma^{'}=$$

and so on?

I thought that these 3 conditions had to be fullfilled to call these elements as a symmetry group

Last edited:
fzero
Homework Helper
Gold Member
Like I said, the requirement that the product of two elements of a set is another element in the set is also a requirement to have a group. When you write

$$(a\cdot b)\cdot c= a\cdot (b\cdot c),$$

you're assuming that $$(a\cdot b)\cdot c$$ is actually in $$\Gamma$$.

However, it's not just that you have to verify closure. It's also that knowing the multiplication table is necessary to verify condition 1 anyway. How could you say that

$$(C_2 \cdot \sigma)\cdot \sigma' = C_2 \cdot ( \sigma\cdot \sigma')$$

if you don't know what $$C_2 \cdot \sigma$$ is equal to?

yes you are right, thank you, but here I meet another problem. I do not know what I will get when for example
$$C_{2}\cdot \sigma=$$ or
$$C_{2}\cdot \sigma^{'}=$$

first I rotate the vase and then mirror reflection....

Last edited:
fzero