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Math Help - Isomorphic or not?

  1. #1
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    Isomorphic or not?

    1. Show that  \mathbb {Q} is not isomorphic to D_4.

    2. Show that no two of D_6 \ , \ T \ , \ A_4 are isomorphic, T = <a,b : a^6 =1, b^2 = a^3, bab^{-1} = a^{-1} >

    3. Show that  \mathbb {H} \   \{ 0 \} is a group under multiplication. H is quaternion algebra.

    My solutions:

    1. The set of rational number is commutative, but D_4 is not, so would that be the reason?

    2. I want to say that T contains element with the order 6 while A_4 does not, would that be right?

    3. For this one I'm just simply lost, would anyone please explain more what H is?

    Thanks.
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  2. #2
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    Quote Originally Posted by tttcomrader View Post

    2. Show that no two of D_6 \ , \ T \ , \ A_4 are isomorphic, T = <a,b : a^6 =1, b^2 = a^3, bab^{-1} = a^{-1} >
    A_4 has no element of order six while D_6 has and so does T.
    While T has a unique subgroup of order six and D_6 has three of order six.

    3. For this one I'm just simply lost, would anyone please explain more what H is?
    Define \mathbb{H} = \mathbb{R}^4 any element of H is written as (a,b,c,d) but we write it as a+bi+cj+dk. Now define i^2=j^2=k^2=-1 and ij=k, jk=i,ki=j,ji=-k,ik=-j,ik=-j. With this you extend this operation as if it was distributive. For example, (1+i+j)(k) = k+ik+jk = k - j + i. Note that (\mathbb{H},\cdot ) is not commutative.
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