Symmetry group

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Homework Statement


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





Homework Equations



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



The Attempt at a Solution


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

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

for example[tex] (C_{2}\cdot \sigma)\cdot e=C_{2}\cdot(\sigma\cdot e)[/tex]???

Homework Statement





Homework Equations





The Attempt at a Solution

 

Answers and Replies

  • #2
fzero
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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 [tex]\Gamma[/tex] is also an element belonging to [tex]\Gamma[/tex]. Therefore, the first thing you want to do is build the multiplication table for the elements of [tex]\Gamma[/tex].

When you know the rules for multiplication, verifying associativity will be fairly simple.
 
  • #3
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you mean I have to calculate
[tex] C_{2}\cdot e=[/tex]
[tex] C_{2}\cdot C_{2}=[/tex]
[tex] C_{2}\cdot \sigma=[/tex]
[tex] C_{2}\cdot \sigma^{'}=[/tex]
[tex] \sigma^{'}\cdot \sigma{'}=[/tex]
[tex] e\cdot e=[/tex]
[tex] e\cdot C_{2}=[/tex]
[tex] e\cdot \sigma=[/tex]
[tex] e\cdot \sigma^{'}=[/tex]

and so on?

I thought that these 3 conditions had to be fullfilled to call these elements as a symmetry group
 
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  • #4
fzero
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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

[tex](a\cdot b)\cdot c= a\cdot (b\cdot c),[/tex]

you're assuming that [tex](a\cdot b)\cdot c[/tex] is actually in [tex]\Gamma[/tex].

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

[tex] (C_2 \cdot \sigma)\cdot \sigma' = C_2 \cdot ( \sigma\cdot \sigma')[/tex]

if you don't know what [tex] C_2 \cdot \sigma[/tex] is equal to?
 
  • #5
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yes you are right, thank you, but here I meet another problem. I do not know what I will get when for example
[tex] C_{2}\cdot \sigma=[/tex] or
[tex] C_{2}\cdot \sigma^{'}=[/tex]

first I rotate the vase and then mirror reflection....
 
Last edited:
  • #6
fzero
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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.
 

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