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I have the following question :

Find $\alpha \in S_4$ such that $\alpha^2=(12)(34)$

I tried to use the method from this : $\alpha=(714)(3925)$ Find $\beta \in S_9$ so $\beta^5=\alpha$

Yet were unsuccessful.

What I did

$lcm(\alpha$)=2 therefore $(\alpha^2)^{2k+1}=\alpha^{4k+2}=\alpha^{2}$.

Therefore using the method from previous topic I get that :

$4k+2=m \rightarrow m=2(mod 4)$ so $m=2$, but $\alpha=[(12)(34)]^2$ is not the answer.

Any ideas?

Thanks

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    Try $\alpha=(1\;\;3\;\;2\;\;4)$. There are only two possible choices of $\alpha$. The other choice is $\alpha=(1\;\;4\;\;2\;\;3)$. Note that $\alpha$ must be an element of order $4$ of $S_4$, and there are only $6$ of them. – 2017-01-28
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    Why $\alpha$ has to be of order $4$? – 2017-01-28
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    Well, $\alpha^4=\left(\alpha^2\right)^2=1$, and $\alpha^2\neq 1$. – 2017-01-28

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Since $\alpha^2$ has order $2$, we can try an element $\alpha$ which has order $4$ in $S_4$, e.g. $\alpha = (1 \; a \; b \; c)$.

Then $$\alpha^2 = (1 b)(a c)$$ I let you find suitable $a,b,c$...

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    But we know that $\alpha^2=(12)(34)$ so $b=2,a=3,c=4??$ but we are looking for $\alpha$ not $\alpha^2$ I don't understand can you expand your answer? – 2017-01-28
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    @JaVaPG : Yes, so you've found $\alpha = (1 3 2 4)$. Does it satisfy the required property? – 2017-01-28
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    But you wrote that $\alpha$ cannot be an element that has order $4$ – 2017-01-28
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    @JaVaPG : Oops, that was a typo, sorry! In general, the order of $a^k$ is $ord(a) / \gcd(ord(a),k)$ – 2017-01-28
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    I know following formula we we are looking in this case for $k=2$ so we have $\frac{ord(\alpha)}{gcd(ord(\alpha),2)}=ord(\alpha^2)=2$ But how do we know $ord(\alpha)$ from this? – 2017-01-28
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    @JaVaPG : from $ord(a) = 2\gcd(ord(a),2)$ you get that $ord(a)$ is even, so $\gcd(ord(a), 2) = 2$ and finally $ord(a)=2 \cdot 2 = 4$. You can also prove it directly, as Barominovski mentioned in the comments above. – 2017-01-28
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    Oh, I see now that only for $ord(\alpha)=4$ the following is true $\frac{ord(\alpha)}{gcd(ord(\alpha),2)}=ord(\alpha^2)=2$ right? – 2017-01-28
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    @JaVaPG : yes this is right. Is my answer OK, now? – 2017-01-28
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    Notice that an element of order $n$ in $S_n$ doesn't have to be a $n$-cycle, for instance $(1 \; 2 \; 3)(4 \; 5) \in S_6$. But for $S_4$ this holds. – 2017-01-29