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

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*}

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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

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...