Thread: Isomorphic or not?

1. 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 = $

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.

2. Originally Posted by tttcomrader

2. Show that no two of $D_6 \ , \ T \ , \ A_4$ are isomorphic, $T = $
$A_4$ 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.