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

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Chris L T521

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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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Chris L T521

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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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