How do symmetry operations create a symmetry group for a Greek vase?

[rayman123]

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 <br /> 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 correctlyC_{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{&#039;}, e]

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

 

[fzero]

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.

 

[rayman123]

you mean I have to calculate
C_{2}\cdot e=
C_{2}\cdot C_{2}=
C_{2}\cdot \sigma=
C_{2}\cdot \sigma^{&#039;}=
\sigma^{&#039;}\cdot \sigma{&#039;}=
e\cdot e=
e\cdot C_{2}=
e\cdot \sigma=
e\cdot \sigma^{&#039;}=

and so on?

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

 

[fzero]

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&#039; = C_2 \cdot ( \sigma\cdot \sigma&#039;)

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

 

[rayman123]

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^{&#039;}=

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

 

[fzero]

You'll want to draw some pictures to work it out. Some of those combinations will just give the identity, others are equivalent to a reflection.