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Problem of the Week #49 - March 4th, 2013

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Chris L T521

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Jan 26, 2012
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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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Remember to read the POTW submission guidelines to find out how to submit your answers!
 
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Chris L T521

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Jan 26, 2012
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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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