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Math Help - Group of small order

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    Question Group of small order

    Let G be a non-Abelian group with eight elements. Show that G has an element a say of order 4. Let b be an element of G which is not e, a, a^2 or a^3. By considering the possible values of b^2 and of ba and ab, show that G is isomorphic either to the dihedral group ot to the quaternion group.
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    Quote Originally Posted by r7iris View Post
    Let G be a non-Abelian group with eight elements. Show that G has an element a say of order 4. Let b be an element of G which is not e, a, a^2 or a^3. By considering the possible values of b^2 and of ba and ab, show that G is isomorphic either to the dihedral group ot to the quaternion group.
    Let us do a slightly tougher problem.

    "Determine all groups of order 8 up to isomorphism".

    Now, if G is abelian the problem is extremely easy by the powerful fundamental theorem of finitely generated abelian groups. In which case it is either:
    \mathbb{Z}_8 \ \ \ \mathbb{Z}_2\times \mathbb{Z}_4 \ \ \ \mathbb{Z}_2\times \mathbb{Z}_2\times \mathbb{Z}_2.

    The hard part is doing the non-abelian case. By Lagrange's theorem the order of an element divides the order of the group, in which case 2,4,8 are the possible orders. Now it cannot be 8 for that will imply the group is cyclic and hence abelian. We show it cannot be have all elements of order 2. Because then (ab)^2 = abab=1\implies ab=ba a contradiction. Let a have order 4. And consider the subgroup \left< a \right> of order 4. And let b\not \in \left< a \right> which means the cosets \left< a \right> \mbox{ and }b\left< a \right> contain all the elements of G. This shows that a,b generate the group G. Since (G:\left< a \right>)=2 it means \left< a \right> \triangleleft G and the factor group G/\left<a\right>\simeq \mathbb{Z}_2. The last sentence implies that b^2 \in \left< a\right>. It cannot be that b^2 = a \mbox{ or }b^2=a^3 for that will imply that b has order 8, which cannot be since G is non-abelian group. Thus, there are two possibilities: b^2 = 1 \mbox{ or }b^2 = a. We mentioned that \left< a \right> \triangleleft G so bab^{-1}\in \left< a\right>. So, bab^{-1}=1\mbox{ or }a\mbox{ or }a^2\mbox{ or }a^3. If bab^{-1}=1\implies a=1 a contradiction. If bab^{-1}=a^2 \implies ba^2b^{-1}=1\implies a^2 = 1 a contradiction. If bab^{-1}=a\implies ab=ba a contradiction. Thus, bab^{-1}=a^3\implies ba=a^3b. All this work leads to two group presentations:
    (a,b : a^4 = 1, b^2 = 1, ba=a^3b)
    (a,b : a^4=1 , b^2 = a^2, ba=a^3b)
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