# Is the group of upper triangular matrices isomorphic to $\mathbb{R}^3$?

Let $$G = \left\{\begin{bmatrix} 1 & x & y \\ 0 & 1 & z \\ 0 & 0 & 1 \end{bmatrix} : x,y,z\in\mathbb{R}\right\}$$

I have shown that the set $G$ with the operation of matrix multiplication is a non-abelian subgroup of $SL(3, \mathbb R)$.

Now, is $G$ isomorphic to the group comprising $\mathbb R^3=\{(x,y,z):x,y,z$ are in $\mathbb R\}$ with the operation of vector addition.

I know I have to show the map is a homomorphism and that it is bijective. I have done the process successfully with simpler groups but don't know what to do for a matrix and vector.

• You already shown that it is not abelian, so clearly it cannot be isomorphic to the given group. – Tobias Kildetoft Nov 10 '16 at 12:28
• See the extensive article on the Heisenberg group. – Dietrich Burde Nov 10 '16 at 14:30
• I changed your title. Isomorphism is a relation between groups. You can't "show a group is isomorphic" any more than you can "show five is less than." – Matthew Leingang Nov 10 '16 at 14:49

If $G$ and $G'$ are isomorphic, any group-theoretic property of $G$ must also hold in $G'$. By “group-theoretic property” I mean any property of the group which depend only on group axioms. Abelianness is such a property. Thus, if $G$ and $G'$ are isomorphic, they are either both abelian or both nonabelian.
Let $G'$ be the group of $\mathbb{R}^3$ under addition. Your parametrization of $G$ shows that there are plenty of bijections between them. But you've also shown that $G$ is nonabelian, while $G'$ is clearly abelian. So the groups cannot be isomorphic.