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Consider the following group of matrices with multiplication:

$$I = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix},\ A = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}, \ B = \begin{pmatrix} 0 & 1 \\ -1 & -1 \end{pmatrix}, \\ C = \begin{pmatrix} -1 & -1 \\ 0 & 1 \end{pmatrix}, \ D = \begin{pmatrix} -1 & -1 \\ 1 & 0 \end{pmatrix}, \ K = \begin{pmatrix} 1 & 0 \\ -1 & -1 \end{pmatrix}$$

I have to determine if this group is isomorphic to $S_3$. The two groups look pretty similar: they both divide into two halves: one is cyclic, and the other consists of elements that are their own inverse. I have been trying for a while to match up their group tables, but there's always something that doesn't fit.

Is there an isomorphism between these groups, and if there isn't, how can I prove it?

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6  
How many isomorphy classes of groups of order $6$ are there? (Hint: it's a very small positive integer.) –  Daniel Fischer Jul 7 '13 at 14:54
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3 Answers

up vote 4 down vote accepted

@DanielFischer has given a very clear and satisfactory answer in his comment.

For an explicit one, consider that $A^2 = I$, $B^3 = I$, $A B A B = I$, so $A, B$ satisfy the standard relations that define $S_{3}$.

Note also $$ B^2 = D, A B = C, A B^2 = K, $$ so $\langle A, B \rangle = \{I, A, B, C, D, K \}$.

This also allows you to write down explicitly an isomorphism as $A \mapsto (12), B \mapsto (123)$.

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+1. Yes. Honestly, I wanted to remove myself cause all I wanted to say was being said very completely by you. Thanks Andreas. –  B. S. Jul 7 '13 at 16:01
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If you are familiar to $S_3$'s presentation: $$S_3=\langle a,b\mid a^2=b^3=(ab)^2=1\rangle=\{e,a, b, ab, b^2, ba \}$$ then you see that $$A^2=I,~~B^3=I, B^2=D,...$$

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+1, we argumented very similarly. –  Andreas Caranti Jul 7 '13 at 15:54
    
@amWhy: my dear friend, I had a very exhausting day. Now, I am out, but maybe for hours later I will be on. Sorry and forgive me. –  B. S. Jul 8 '13 at 15:38
    
Oh, of course, dear friend. Life's demands come first! –  amWhy Jul 8 '13 at 15:43
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A good way to learn things about a given group is to let it act. Let this group $G$ act on the complex projective line (any projective line over a field would work) $$ \mathbb{P}^1(\mathbb{C})=\{[x:y]\,;\,(x,y)\in\mathbb{C}^2\setminus\{(0,0)\}\} $$ by $g\cdot[x:y]:=[g(x,y)]$, the natural action of $GL(2,\mathbb{C})$ on $\mathbb{P}^1(\mathbb{C})$. With the three particular points $$ 1=[1:0]\quad 2=[0:1]\quad 3=[1:-1] $$ you can easily check that $G$ acts faithfully on $\{1,2,3\}$. With this identification: $$ I=Id\;\; A=(1,2)\; B=(1,2,3)\; C=(2,3)\; D=(1,3,2)\; K=(1,3) $$ whence $G\simeq S_3$.

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Nicely done. It actually works also in characteristic $2$, it seems, where the group acts directly on the three non-zero points of the two-dimensional vector space. –  Andreas Caranti Jul 7 '13 at 16:19
    
@AndreasCaranti You're right, thanks for the note. –  1015 Jul 7 '13 at 16:28
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