Quaternions isomormphism question

nhk

Apr 2010
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Let G be the group {±1, ±i, ±j, ±k} where i^2=j^2=k^2=–1, –i=(–1)i, 1^2=(–1)2=1, ij=–ji=k, jk=–kj=i, and ki=–ik=j. (Quaternions group multiplication). Prove that G is isomorphic to
Q = <a, b|a^4=e, b^2=a^2, and ab=b(a^3)>.
 

NonCommAlg

MHF Hall of Honor
May 2008
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Let G be the group {±1, ±i, ±j, ±k} where i^2=j^2=k^2=–1, –i=(–1)i, 1^2=(–1)2=1, ij=–ji=k, jk=–kj=i, and ki=–ik=j. (Quaternions group multiplication). Prove that G is isomorphic to
Q = <a, b|a^4=e, b^2=a^2, and ab=b(a^3)>.
first of all, since \(\displaystyle ab=ba^3\) and \(\displaystyle a^4=b^4=1\), every element of \(\displaystyle Q\) can be written as \(\displaystyle a^rb^s,\) for some \(\displaystyle 0 \leq r \leq 3, \ 0 \leq s \leq 3.\) now use the relations again to find all 8 distinct element of \(\displaystyle Q.\)

then define a map from \(\displaystyle Q\) to \(\displaystyle G\) to send \(\displaystyle a\) to \(\displaystyle i\) and \(\displaystyle b\) to \(\displaystyle j\).
 
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nhk

Apr 2010
43
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I am only getting 7 distinct elements ( even though I know that Q has order 8), the elements that I am getting are: a,a^2, a^3, e, b, b^3, and ab. Since ab= ba^3, I know that ab=ba^3=b^3a (since a^2=b^2). So since ab=b^3a, than b^-2ab=ba which implies that b^2ab=ba and that a^3b=ba=ab^3. So I can't seem to get an eigth element, since a^2b^2=e, a^3b^2=a, a^2b^3=b.
 

NonCommAlg

MHF Hall of Honor
May 2008
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I am only getting 7 distinct elements ( even though I know that Q has order 8), the elements that I am getting are: a,a^2, a^3, e, b, b^3, and ab. Since ab= ba^3, I know that ab=ba^3=b^3a (since a^2=b^2). So since ab=b^3a, than b^-2ab=ba which implies that b^2ab=ba and that a^3b=ba=ab^3. So I can't seem to get an eigth element, since a^2b^2=e, a^3b^2=a, a^2b^3=b.
well, yuo just said it yourself, the eigth element is \(\displaystyle a^3b=ba.\)
 
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nhk

Apr 2010
43
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Oh man, I can't believe I just made that stupid of a response, thanks for your help.