# 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

Tobias has nailed it. It's important to understand what isomorphism between groups means.

Yes, two groups are isomorphic if there is a bijective homomorphism between them. But more importantly, it means they are indistinguishable as far as groups go. Poincaré said that mathematics is the art of calling different things by the same name. Isomorphic groups get the same “name” in algebra.

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.

One way to show that two groups are not isomorphic is to find a group-theoretic property that one holds and the other does not. It's easier and cleaner than assuming there is an isomorphism and trying to derive a contradiction.

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.