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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.

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2 Answers 2

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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.

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  • $\begingroup$ Okay, Thanks! The question is also asking about center of G. I think the center of G is {0, 1, a,b,c}, Is it correct? $\endgroup$
    – Tanay
    Apr 19, 2020 at 11:33
  • $\begingroup$ sir your answer is very helpful to me and I upvoted it. But unfortunately I'm having less than 15 reputation for that. Please say if I'm right or wrong regarding the center of G sir. $\endgroup$
    – Tanay
    Apr 19, 2020 at 11:36
  • $\begingroup$ sir how do I accept it? Please tell the steps $\endgroup$
    – Tanay
    Apr 19, 2020 at 11:40
  • $\begingroup$ Thank You so much for your help! $\endgroup$
    – Tanay
    Apr 19, 2020 at 11:41
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The question must have meant to say non-Abelian. Incidentally, this is called the Heisenberg group.

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    $\begingroup$ Here it is the Heisenberg group over $\Bbb Q$. The coefficients can be an arbitrary commutative ring with identity (see the link). $\endgroup$ Apr 19, 2020 at 8:19

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