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We have to prove that $H=\{i,(12),(34),(12)(34)\}$ forms a non-cyclic subgroup of $S_4$.

It is seen that there is no element of order $4$ in $H$. So, $H$ is non-cyclic. But How can I show $H$ is a subgroup of $G$. Is there any shortcut method?

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  • $\begingroup$ Very closely related. This group has been handled so many times. Did you search the site? $\endgroup$ – Jyrki Lahtonen May 25 at 11:57
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Labeling, $a=(12)$, $b=(34)$. You have $a$ and $b$ commute ($ab=ba$) and $a^2=b^2=i$. From the previous facts, it's not difficult to see that $H=\{i,a,b,ab\}$ is closed under the group operation, so it is a subgroup. In fact $H$ is isomorphic to $\mathbb{Z}/2\mathbb{Z}\times\mathbb{Z}/2\mathbb{Z}$ by the map given by $a\mapsto (1,0)$ and $b\mapsto (0,1)$.

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  • $\begingroup$ product of disjoint cycles is not commutative in general? right?@JulianMejia $\endgroup$ – MKS May 25 at 11:09
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    $\begingroup$ If the cycles are disjoint then their product is always commutative. But of course a product of permutations in general might be not commutative. $\endgroup$ – Mark May 25 at 11:10
  • $\begingroup$ If $a=(x_1,x_2,\dots,x_n)$ and $b=(y_1,y_2,\dots,y_m)$ are disjoint cycles (i.e. $x$'s and $y$'s don't intersect). Then you have that $ab=ba$. Because $a$ permutes the $x's$ and $b$ is permuting the $y$'s. It doesn't matter if you do first $a$ and then $b$ or viceversa, you get the same thing since $x$'s and $y$'s are independent. $\endgroup$ – Julian Mejia May 25 at 11:16
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Yes. It's enough to show it is closed (it is almost obvious here), since $H$ is finite; see here.

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