Quaternions isomormphism question

nhk

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
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 $$\displaystyle j$$.

nhk

nhk

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
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 $$\displaystyle a^3b=ba.$$

nhk

nhk

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