# Commutative Property under Matrix multiplication

I am asked to show that the set of matrices $$G=\left\{\begin{bmatrix}1&a&b\\0&1&c\\0&0&1\end{bmatrix}:a,b,c\in\mathbb Q\right\}$$ form an abelian group wrt matrix multiplication. (Assume that matrix multiplication is associative).

I know that for $$G$$ to form an abelian group under matrix multiplication,

1. Matrix multiplication in $$G$$ should be associative.
2. Existence of identity element in matrix multiplication.
3. Existence of inverse element in matrix multiplication.
4. Matrix multiplication in $$G$$ should be commutative.

For $$1$$, it is already given that matrix multiplication is associative.

For $$2$$, I have found the identity element $$a = b = c = 0$$.

For $$3$$, I have similarly found the inverse element.

But for $$4$$, I am unable to prove that $$AB = BA$$ for all $$A,B\in G$$ containing terms a1,b1,c1,a2,b2,c2 since a1c2 is not equal to a2c1.

How do show that $$G$$ is an abelian group under matrix multiplication? Please help.

You can't prove commutativity because it's false. Let $$A=\begin{bmatrix}1&1&0\\0&1&0\\0&0&1\end{bmatrix}\qquad B=\begin{bmatrix}1&0&0\\0&1&1\\0&0&1\end{bmatrix}$$ Then $$AB=\begin{bmatrix}1&1&1\\0&1&1\\0&0&1\end{bmatrix}\ne\begin{bmatrix}1&1&0\\0&1&1\\0&0&1\end{bmatrix}=BA$$ $$G$$ is still a group under matrix multiplication, but not an abelian group.
• Here it is the Heisenberg group over $\Bbb Q$. The coefficients can be an arbitrary commutative ring with identity (see the link). Apr 19, 2020 at 8:19