# Prove that $H=\{i,(12),(34),(12)(34)\}$ forms a non-cyclic subgroup of $S_4$.

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?

• Very closely related. This group has been handled so many times. Did you search the site? – Jyrki Lahtonen May 25 at 11:57

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 $$b\mapsto (0,1)$$.
• 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. – Julian Mejia May 25 at 11:16
Yes. It's enough to show it is closed (it is almost obvious here), since $$H$$ is finite; see here.