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I am asked to find an isomorphism from the group $G = 1,i,-1,-i$ to the group $H = \begin{bmatrix} 1 & 0 \\[0.3em] 0 & 1 \end{bmatrix} , \begin{bmatrix} i & 0 \\[0.3em] 0 & -i \end{bmatrix} \begin{bmatrix} -i & 0 \\[0.3em] 0 & i \end{bmatrix} \begin{bmatrix} -1 & 0 \\[0.3em] 0 & -1 \end{bmatrix} $

it is easy to see that $f(1) = \begin{bmatrix} -1 & 0 \\[0.3em] 0 & -1 \end{bmatrix}$ and $f(-1) = \begin{bmatrix} -1 & 0 \\[0.3em] 0 & -1 \end{bmatrix}$ because identities must map to identities and also inverses must map to inverses and we observe that $-1$ is its own inverse same like $\begin{bmatrix} -1 & 0 \\[0.3em] 0 & -1 \end{bmatrix}$ and now we have two elements left to map to we can map $i$ to $\begin{bmatrix} -i & 0 \\[0.3em] 0 & i \end{bmatrix}$ and $-i$ to $\begin{bmatrix} i & 0 \\[0.3em] 0 & -i \end{bmatrix}$ ,this is an isomorphims but how to prove that $f(xy) = f(x)f(y)$

alkabary
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  • How about trying to multiply the matrices together and see if it gives the wanted result? – Patrick Da Silva Oct 29 '14 at 23:52
  • yes but this will be too long though, is there a concrete way to do it – alkabary Oct 29 '14 at 23:53
  • If you don't see anything or any fancy trick, you should at least try computing some stuff to see if some useful identities show up. Then you can see what's going on with the computations in front of you and deduce the fancy tricks. With experience you learn to see the fancy tricks before starting to compute, but experience doesn't just magically show up. – Patrick Da Silva Oct 29 '14 at 23:58

2 Answers2

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Note that $G$ consists of all powers of $i$, and $H$ consists of all powers of $$A=\pmatrix{i&0\cr0&-i\cr}\ .$$ So use the map $$f(i^k)=A^k$$ for all $k$. Then you have $$f(xy)=f(i^ki^l)=f(i^{k+l})=A^{k+l}=A^kA^l=f(i^k)f(i^l)=f(x)f(y)\ .$$

David
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Quaternions can be represented as $2\times 2$ complex matrices: $$ a + bi + cj + dk \ \equiv \ \begin{bmatrix}a+bi & c+di \\ -c+di & a-bi \end{bmatrix} $$ This representation is a ring isomorphism.

Under this representation, complex numbers are represented by $$ a + bi \ \equiv \ \begin{bmatrix}a+bi & 0 \\ 0 & a-bi \end{bmatrix} $$ This is the isomorphism mentioned in the question follows from this.

lhf
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