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Is the group $H$ consisting of matrices of the form $ \left( {\begin{array}{cc} 1 & n \\ 0 & 1 \\ \end{array} } \right) $ cyclic, where $n \in \mathbb{Z}$? If not, how would you show this?

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  • $\begingroup$ did you note that $\left( {\begin{array}{cc}1&1\\0&1\end{array}}\right)\left( {\begin{array}{cc}1&1\\0&1\end{array}}\right)=\left( {\begin{array}{cc}1&2\\0&1\end{array}}\right)$? $\endgroup$
    – janmarqz
    Commented Nov 1, 2015 at 14:05
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    $\begingroup$ Possible duplicate of math.stackexchange.com/questions/1493945/…. $\endgroup$
    – lhf
    Commented Nov 1, 2015 at 14:06
  • $\begingroup$ Why has a question about a 2x2-matrix the tag "infinite-matrices"? $\endgroup$ Commented Jan 28, 2016 at 18:45

5 Answers 5

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The group is indeed cyclic and is generated by the matrix $$ J=\left( {\begin{array}{cc} 1 & 1 \\ 0 & 1 \\ \end{array} } \right) $$

As you have for $n \in \mathbb Z$

$$ J^n =\left( {\begin{array}{cc} 1 & n \\ 0 & 1 \\ \end{array} } \right) $$

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if we define: $$ B=\left( {\begin{array}{cc} 0 & 1 \\ 0 & 0 \\ \end{array} } \right) $$ then $$ A = I+B \\ B^2 = 0 $$ so by the binomial theorem, for $n \ge 0$: $$ A^n =(I+B)^n = I+nB $$ note also that: $$ (I+B)(I-B) = I-B^2=I $$ so $$ A^{-1}=I-B \\ A^{-n}=(I-B)^n =I-nB $$

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Hint:

Compute the product $\begin{pmatrix}1&n\\0&1\end{pmatrix}\begin{pmatrix}1&p\\0&1\end{pmatrix}$.

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Observe that $\left( {\begin{array}{cc}1&1\\0&1\end{array}}\right)^m=\left( {\begin{array}{cc}1&m\\0&1\end{array}}\right)$ for each $m\in\Bbb Z$.

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Here's to get you started. I propose that $H$ is cyclic with generator $A=\left( {\begin{array}{cc} 1 & 1 \\ 0 & 1 \\ \end{array} } \right)$.

Why is this so? Try calculating $A^2$, and in general, you can prove by induction that $A^n = \left( {\begin{array}{cc} 1 & n \\ 0 & 1 \\ \end{array} } \right)$.

Don't forget the matrices for negative $n$. What is $A^{-1}$? How do you get $\left( {\begin{array}{cc} 1 & -n \\ 0 & 1 \\ \end{array} } \right)$ from $A$?

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  • $\begingroup$ How exactly would you consider instances where $n<0$? I mean, a matrix $M$ to the power of $-1$ would give $\frac{1}{M}$, wouldn't it? $\endgroup$
    – M Smith
    Commented Nov 1, 2015 at 17:21
  • $\begingroup$ @user2910074 The matrix $M^{-1}$ is the inverse of $M$. I think you will find that this inverse is in you group of matrices for $n=-1$. Also, $M^{-2}$ is equal to the inverse squared, and so on, i.e. $M^{-k} = (M^{-1})^k$. $\endgroup$
    – Mankind
    Commented Nov 2, 2015 at 13:04

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