# Problem of the Week #36 - February 4th, 2013

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#### Chris L T521

##### Well-known member
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Here's this week's problem.

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Problem: Show that $\textrm{GL}_2(\mathbb{Z}/2\mathbb{Z})$ is isomorphic to $\text{Sym}(3)$ (also denoted as $S_3$).

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Remember to read the POTW submission guidelines to find out how to submit your answers!

#### Chris L T521

##### Well-known member
Staff member
This week's problem was correctly answered by jakncoke. You can find his solution here:

I will use the following theorem in this proof, every group of order 2p for p > 2 is isomorphic to either $\mathbb{Z_{2p}}$ or $\mathbb{D_{p}}$ (I will give a proof of this if required)

$\mathbb{Z}/2\mathbb{Z} \cong \mathbb{Z_{2}}$ so $GL_2(\mathbb{Z_{2}})$ has 6 elements. For if $A = \begin{bmatrix}a_1 & a_2 \\ a_3 & a_4 \end{bmatrix}$ $a_i \in \mathbb{Z_{2}}$ then Det(A) = 1 only when one of the diagonals has 1,1 as its entries and the other 2 entries can be either (0,1),(1,0), or (0,0).
So 2 * 3 = 6 possible matricies with 1 as its determinant.

So $GL_2(\mathbb{Z_{2}})\cong Z_6$ or $\cong D_3$.

Note that $Z_6$ has only 1 element of order 2. namely $3 \in Z_6$ where as

$\begin{bmatrix}1 & 0 \\ 1 & 1\end{bmatrix}$ and $\begin{bmatrix}1 & 1 \\ 0 & 1\end{bmatrix}$ both elements of $GL(\mathbb{Z_2})$ have order 2.

So clearly $GL_2(\mathbb{Z_{2}}) \not \cong \mathbb{Z_{6}}$ so $GL_2(\mathbb{Z_{2}}) \cong D_{6}$

Now again Sym(3) has 3! = 3*2 elements, and so $Sym(3) \cong Z_{6}$ or $\cong D_{3}$. again Sym(3) has more than 1 element of order 2.
namely (12)(3), (13)(2), ... , so $Sym(3) \cong D_{3}$.

so $Sym(3) \cong D_{3} \cong GL_2(\mathbb{Z_{2}}) \cong GL_2(\mathbb{Z}/2\mathbb{Z})$

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