Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Find an isomorphism from the octic group $G$ to the group $G'$:

$G = \{e, s, s^2, s^3, b, g, d, t\}$

$$e = (1)\quad s = (1234)\quad s^2= (13)(24)\quad s^3= (1432)\quad b = (14)(23)\quad g = (24)\quad d = (12)(34)\quad t = (13)$$

$G' = \{I_2, R, R^2, R^3, H, D, V, T\}$

$$\begin{align*}I_2&=\begin{pmatrix}1 & 0\\ 0& 1 \end{pmatrix}& R&=\begin{pmatrix} 0& -1\\1 &0 \end{pmatrix}& R^2&=\begin{pmatrix}-1 & 0\\ 0&-1 \end{pmatrix}& R^3&=\begin{pmatrix} 0& 1\\-1 & 0\end{pmatrix}\\\\ H&=\begin{pmatrix} 1& 0\\0 & -1 \end{pmatrix}& V&=\begin{pmatrix} -1& 0\\0 & 1 \end{pmatrix}& D&=\begin{pmatrix} 0& 1\\1& 0\end{pmatrix}& T&=\begin{pmatrix} 0& -1\\ -1& 0\end{pmatrix}\end{align*}$$

share|improve this question
    
This looks like homework. Please consider tagging it as such. –  Aryabhata Oct 22 '10 at 1:33
    
no the homework didn't work out the groups... I did –  Jessica Oct 22 '10 at 1:37
    
I was given a a square and had to figure out the responding rigid motions. thus G was found by me. It's an ongoing problem I am trying to figure out. –  Jessica Oct 22 '10 at 1:39
    
@Jessica: learn the concept of homomorphism first. It's not enough to find a bijection between two groups to prove they're isomorphic; you need to show that the mapping you establish does preserve the group operation, that is, if a' and b' are the elements in G' that correspond with (via that mapping) a and b, then a . b = a' .' b', where . is the operation in G and .' is the operation in G'. –  Weltschmerz Oct 22 '10 at 1:53
    
so an isomorphism must take the identity to the identity and preserves the order of elements. So I should be looking at s and R both having order 4. Wait does that make sense? All R, H, D, V... have order 4. ugh –  Jessica Oct 22 '10 at 2:03

1 Answer 1

Jessica: Preserving the order of all elements isn't sufficient; you have to show that it preserves the group operation.

Here's a hint though. $G^\prime$ is the set of symmetries of a square, right? Try labelling the square's corners. Then each symmetry is just a permutation of its corners...

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.