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Consider the set of matrices G = {[1, a, b; 0, 1, c; 0, 0, 1] : a, b, c belong to Q (set of rational numbers)} Given that matrix multiplication is associative, find the center of G.

Since we know that the center of a group, Z(G) = {a belong to G: ax=xa for all x belong to G} Hence Z(G) = {0, 1, a, b, c} for all a, b, c belong to Q

Is it correct?

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    $\begingroup$ It's wrong: the elements of $G$ are matrices, so $Z(G)$ should consist of matrices. $\endgroup$ Commented Apr 19, 2020 at 11:59
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    $\begingroup$ If you want to make it easier for people to read your question, and therefore to have a larger pool of potential answerers, please typeset your question using mathjax. You can even do matrices. $\endgroup$ Commented Apr 19, 2020 at 13:24
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    $\begingroup$ @Tanay Yes, that's much better. $\endgroup$ Commented Apr 19, 2020 at 17:15
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    $\begingroup$ This is the Heisenberg group, which you know from here. It's center is generated by the matrix with $b=1$ and $a=c=0$. $\endgroup$ Commented Apr 19, 2020 at 18:33
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    $\begingroup$ @DietrichBurde that only generates the matrices with integer entries. $\endgroup$ Commented Apr 19, 2020 at 19:47

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Hint: Just multiply two such generic matrices: $$\pmatrix{1&a&b\\&1&c\\&&1}\cdot \pmatrix{1&x&y\\&1&z\\&&1}\ =\ \pmatrix{1& x+a& y+az+b\\&1&z+c\\&&1}$$ If we exchange the variables $a,b,c$ with $x,y,z$, hence the order of the two matrices, we arrive to $$\pmatrix{1& a+x& b+xc+y\\&1&c+z\\&&1}\,. $$

So, they commute iff $az=xc$, and for fixed $a,b,c$, this happens for all $x,y,z$ only if $a=c=0$.

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