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

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

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

My solutions:

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

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

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

Thanks.