# Isomorphic or not?

• Sep 13th 2008, 01:57 PM
Isomorphic or not?
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.
• Sep 13th 2008, 04:41 PM
ThePerfectHacker
Quote:

2. Show that no two of $\displaystyle D_6 \ , \ T \ , \ A_4$ are isomorphic, $\displaystyle T = <a,b : a^6 =1, b^2 = a^3, bab^{-1} = a^{-1} >$
$\displaystyle A_4$ has no element of order six while $\displaystyle D_6$ has and so does $\displaystyle T$.
While $\displaystyle T$ has a unique subgroup of order six and $\displaystyle D_6$ has three of order six.
Define $\displaystyle \mathbb{H} = \mathbb{R}^4$ any element of $\displaystyle H$ is written as $\displaystyle (a,b,c,d)$ but we write it as $\displaystyle a+bi+cj+dk$. Now define $\displaystyle i^2=j^2=k^2=-1$ and $\displaystyle ij=k, jk=i,ki=j,ji=-k,ik=-j,ik=-j$. With this you extend this operation as if it was distributive. For example, $\displaystyle (1+i+j)(k) = k+ik+jk = k - j + i$. Note that $\displaystyle (\mathbb{H},\cdot )$ is not commutative.