text: # Problem of the Week #49 - March 4th, 2013
  
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  #### Chris L T521
  
  ##### Well-known member
  Staff member
  Thanks to those who participated in last week's POTW!! Here's this week's problem!
  
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  Problem: Let $\displaystyle H=\left\{\left.\begin{bmatrix} 1 & a & b\\ 0 & 1 & 0\\ 0 & 0 & 1\end{bmatrix}\right| \,a,b\in\mathbb{Z}_3\right\}$. Show that $H$ is an abelian group of order 9. Is $H\cong \mathbb{Z}_9$ or is $H\cong \mathbb{Z}_3\oplus\mathbb{Z}_3$?
  
  (In this problem, $\cong$ means "isomorphic to".)
  
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  #### Chris L T521
  
  ##### Well-known member
  Staff member
  This week's question was correctly answered by jakncoke and Sudharaka. You can find Sudharaka's solution here:
  
  Take any two elements $$\begin{bmatrix} 1 & a_1 & b_1\\ 0 & 1 & 0\\ 0 & 0 & 1\end{bmatrix}, \begin{bmatrix} 1 & a_2 & b_2\\ 0 & 1 & 0\\ 0 & 0 & 1\end{bmatrix}\in H$$. Then,
  
  $\begin{bmatrix} 1 & a_1 & b_1\\ 0 & 1 & 0\\ 0 & 0 & 1\end{bmatrix}.\begin{bmatrix} 1 & a_2 & b_2\\ 0 & 1 & 0\\ 0 & 0 & 1\end{bmatrix}=\begin{bmatrix} 1 & a_1+a_2 & b_1+b_2\\ 0 & 1 & 0\\ 0 & 0 & 1\end{bmatrix}$
  
  Since, $$a_1+a_2$$ and $$b_1+b_2$$ should be congruent to $$0, 1$$ or $$2$$ under modulo $$3$$ we have, $$a_1+a_2,\,b_1+b_2\in\mathbb{Z}_3$$. Therefore,
  
  $\begin{bmatrix} 1 & a_1 & b_1\\ 0 & 1 & 0\\ 0 & 0 & 1\end{bmatrix}.\begin{bmatrix} 1 & a_2 & b_2\\ 0 & 1 & 0\\ 0 & 0 & 1\end{bmatrix}\in H$
  
  Hence $$H$$ is closed under matrix multiplication.
  
  Since matrix multiplication is associative $$H$$ under matrix multiplication also satisfies associativity.
  
  Note that for each $$\begin{bmatrix} 1 & a & b\\ 0 & 1 & 0\\ 0 & 0 & 1\end{bmatrix}\in H$$ we get,
  
  $\begin{bmatrix} 1 & a & b\\ 0 & 1 & 0\\ 0 & 0 & 1\end{bmatrix}.\begin{bmatrix} 1 & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & 1\end{bmatrix}=\begin{bmatrix} 1 & a_1 & b\\ 0 & 1 & 0\\ 0 & 0 & 1\end{bmatrix}$
  
  Therefore the identity element of $$H$$ is, $$\begin{bmatrix} 1 & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & 1\end{bmatrix}$$.
  
  Also for each, $$\begin{bmatrix} 1 & a & b\\ 0 & 1 & 0\\ 0 & 0 & 1\end{bmatrix}\in H$$ we have,
  
  $\begin{bmatrix} 1 & a & b\\ 0 & 1 & 0\\ 0 & 0 & 1\end{bmatrix}.\begin{bmatrix} 1 & -a & -b\\ 0 & 1 & 0\\ 0 & 0 & 1\end{bmatrix}=\begin{bmatrix} 1 & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & 1\end{bmatrix}$
  
  Therefore an inverse exists for each element of $$H$$.
  
  Finally,
  
  $\begin{bmatrix} 1 & a_1 & b_1\\ 0 & 1 & 0\\ 0 & 0 & 1\end{bmatrix}.\begin{bmatrix} 1 & a_2 & b_2\\ 0 & 1 & 0\\ 0 & 0 & 1\end{bmatrix}=\begin{bmatrix} 1 & a_1+a_2 & b_1+b_2\\ 0 & 1 & 0\\ 0 & 0 & 1\end{bmatrix}=\begin{bmatrix} 1 & a_2 & b_2\\ 0 & 1 & 0\\ 0 & 0 & 1\end{bmatrix}.\begin{bmatrix} 1 & a_1 & b_1\\ 0 & 1 & 0\\ 0 & 0 & 1\end{bmatrix}$
  
  Hence $$H$$ is Abelian.
  
  Since we have three choices for both $$a$$ and $$b$$ (0, 1 and 2) there are $$3\times 3=9$$ elements in the group $$H$$. By the Fundamental Theorem of Finite Abelian Groups we have $$H\cong \mathbb{Z}_3\oplus\mathbb{Z}_3$$.
  
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explanation: The document contains a math problem and its solution. The problem asks to show that a specific set \( H \) is an abelian group of order 9 and to determine if \( H \) is isomorphic to \( \mathbb{Z}_9 \) or \( \mathbb{Z}_3 \oplus \mathbb{Z}_3 \). The solution provided demonstrates that \( H \) is indeed an abelian group of order 9 and concludes that \( H \cong \mathbb{Z}_3 \oplus \mathbb{Z}_3 \).
distilabel_metadata.content: Evaluate the user given text extract and determine if it contains a math question and optionally and answer to this question.
  
  Output rules:
  
  1. Briefly describe the document and mention if it contains a question or a question and answer.
  2. Finalize the output with: Label: no-question,question,question-and-answer,multiple-question,multiple-question-and-answer
distilabel_metadata.role: system
distilabel_metadata.content: # Problem of the Week #49 - March 4th, 2013
  
  Status
  Not open for further replies.
  
  #### Chris L T521
  
  ##### Well-known member
  Staff member
  Thanks to those who participated in last week's POTW!! Here's this week's problem!
  
  -----
  
  Problem: Let $\displaystyle H=\left\{\left.\begin{bmatrix} 1 & a & b\\ 0 & 1 & 0\\ 0 & 0 & 1\end{bmatrix}\right| \,a,b\in\mathbb{Z}_3\right\}$. Show that $H$ is an abelian group of order 9. Is $H\cong \mathbb{Z}_9$ or is $H\cong \mathbb{Z}_3\oplus\mathbb{Z}_3$?
  
  (In this problem, $\cong$ means "isomorphic to".)
  
  -----
  
  #### Chris L T521
  
  ##### Well-known member
  Staff member
  This week's question was correctly answered by jakncoke and Sudharaka. You can find Sudharaka's solution here:
  
  Take any two elements $$\begin{bmatrix} 1 & a_1 & b_1\\ 0 & 1 & 0\\ 0 & 0 & 1\end{bmatrix}, \begin{bmatrix} 1 & a_2 & b_2\\ 0 & 1 & 0\\ 0 & 0 & 1\end{bmatrix}\in H$$. Then,
  
  $\begin{bmatrix} 1 & a_1 & b_1\\ 0 & 1 & 0\\ 0 & 0 & 1\end{bmatrix}.\begin{bmatrix} 1 & a_2 & b_2\\ 0 & 1 & 0\\ 0 & 0 & 1\end{bmatrix}=\begin{bmatrix} 1 & a_1+a_2 & b_1+b_2\\ 0 & 1 & 0\\ 0 & 0 & 1\end{bmatrix}$
  
  Since, $$a_1+a_2$$ and $$b_1+b_2$$ should be congruent to $$0, 1$$ or $$2$$ under modulo $$3$$ we have, $$a_1+a_2,\,b_1+b_2\in\mathbb{Z}_3$$. Therefore,
  
  $\begin{bmatrix} 1 & a_1 & b_1\\ 0 & 1 & 0\\ 0 & 0 & 1\end{bmatrix}.\begin{bmatrix} 1 & a_2 & b_2\\ 0 & 1 & 0\\ 0 & 0 & 1\end{bmatrix}\in H$
  
  Hence $$H$$ is closed under matrix multiplication.
  
  Since matrix multiplication is associative $$H$$ under matrix multiplication also satisfies associativity.
  
  Note that for each $$\begin{bmatrix} 1 & a & b\\ 0 & 1 & 0\\ 0 & 0 & 1\end{bmatrix}\in H$$ we get,
  
  $\begin{bmatrix} 1 & a & b\\ 0 & 1 & 0\\ 0 & 0 & 1\end{bmatrix}.\begin{bmatrix} 1 & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & 1\end{bmatrix}=\begin{bmatrix} 1 & a_1 & b\\ 0 & 1 & 0\\ 0 & 0 & 1\end{bmatrix}$
  
  Therefore the identity element of $$H$$ is, $$\begin{bmatrix} 1 & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & 1\end{bmatrix}$$.
  
  Also for each, $$\begin{bmatrix} 1 & a & b\\ 0 & 1 & 0\\ 0 & 0 & 1\end{bmatrix}\in H$$ we have,
  
  $\begin{bmatrix} 1 & a & b\\ 0 & 1 & 0\\ 0 & 0 & 1\end{bmatrix}.\begin{bmatrix} 1 & -a & -b\\ 0 & 1 & 0\\ 0 & 0 & 1\end{bmatrix}=\begin{bmatrix} 1 & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & 1\end{bmatrix}$
  
  Therefore an inverse exists for each element of $$H$$.
  
  Finally,
  
  $\begin{bmatrix} 1 & a_1 & b_1\\ 0 & 1 & 0\\ 0 & 0 & 1\end{bmatrix}.\begin{bmatrix} 1 & a_2 & b_2\\ 0 & 1 & 0\\ 0 & 0 & 1\end{bmatrix}=\begin{bmatrix} 1 & a_1+a_2 & b_1+b_2\\ 0 & 1 & 0\\ 0 & 0 & 1\end{bmatrix}=\begin{bmatrix} 1 & a_2 & b_2\\ 0 & 1 & 0\\ 0 & 0 & 1\end{bmatrix}.\begin{bmatrix} 1 & a_1 & b_1\\ 0 & 1 & 0\\ 0 & 0 & 1\end{bmatrix}$
  
  Hence $$H$$ is Abelian.
  
  Since we have three choices for both $$a$$ and $$b$$ (0, 1 and 2) there are $$3\times 3=9$$ elements in the group $$H$$. By the Fundamental Theorem of Finite Abelian Groups we have $$H\cong \mathbb{Z}_3\oplus\mathbb{Z}_3$$.
  
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distilabel_metadata.role: user
distilabel_metadata.raw_output_q_a_detector_0: The document contains a math problem and its solution. The problem asks to show that a specific set \( H \) is an abelian group of order 9 and to determine if \( H \) is isomorphic to \( \mathbb{Z}_9 \) or \( \mathbb{Z}_3 \oplus \mathbb{Z}_3 \). The solution provided demonstrates that \( H \) is indeed an abelian group of order 9 and concludes that \( H \cong \mathbb{Z}_3 \oplus \mathbb{Z}_3 \).
  
  Label: question-and-answer