# Suppose that $\alpha$ and $\beta$ in $S_4$, $\alpha (3) = 4$, $\alpha\beta = (3412)$, $\beta\alpha = (3241)$, find $\alpha$ and $\beta$

Suppose that $\alpha,\beta \in S_4, \,\alpha (3) = 4,\, \alpha\beta = (3412), \,\beta\alpha = (3241).\;$ Find $\alpha\,$ and $\, \beta$.

My attempt:

Ok so I know

$\alpha(3) = 4$ and $\alpha(\beta(3))=4$ so $\beta(3)=3$

$\beta(3)=3$ and $\beta(\alpha(1))=3$ so $\alpha(1) = 3$

$\alpha(1) = 3$ and $\alpha(\beta(2))=4$ so $\beta(2)=1$

$\beta(2)=1$ and $\beta(\alpha(4))=1$ so $\alpha(4) =2$

$\alpha(4) = 2$ and $\alpha(\beta(1))=2$ so $\beta(1)=4$

But then I just get back to $\alpha(3) = 4$, and am stuck.

Any help would be appreciated. Thanks.

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What is $i$? Is $\alpha(3)$ equal to $3$ as in the title or $4$ as in the text? – Jim Feb 10 '13 at 19:11
my mistake, $\alpha(3) = 4$ – bobdylan Feb 10 '13 at 19:14
Suggest writing down the cycles for $\alpha$ and $\beta$ - what are you left with, and can you fill the gaps? – Mark Bennet Feb 10 '13 at 19:19

Every permutation is a bijection, including $\alpha$ and $\beta$.

Hence, since you've found $\beta(1) = 4, \beta(2) = 1, \beta (3) = 3,$ that leaves only the possibility that $\beta(4)=2$.

Similarly, $\alpha(1) = 3, \alpha(2) = ?, \alpha(3) = 4, \alpha(4) = 2 \implies \alpha(2) = 1$

So write the cycles: $\alpha, \beta$ and you're done:

$$\alpha: 1\to 3, \;3 \to 4,\; 4\to 2,\; 2 \to 1 \;\implies\; \alpha = (1342)$$ $$\beta: 1 \to 4, \;4 \to 2, \;2 \to 1;\; \,3\to 3 \;\implies \;\beta = (142)(3) = (142)$$

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$\beta$ is a bijection. you already found that $\beta(1)=4,\beta(2)=1,\beta(3)=3$. Thus, $\beta(4)=2$. Now that you find $\beta$ you can get $\alpha=(3412)\beta^{-1}$

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+1: Or, for that matter, the OP could use similar argument to determine $\alpha(2)$, since the other three function values have been determined. – Cameron Buie Feb 10 '13 at 19:39
@Cameron Buie Yes thats less work – Amr Feb 10 '13 at 19:41