Let $G$ be a subgroup of $S_8$ that act on $\Omega = > \{\{1,2\},\{3,4\},\{5,6\},\{7,8\}\}$. Find $|G|$ by using a homomorphism $\Phi:G \rightarrow Perm(\Omega)$ where $\Phi(g) = > \sigma_g$, a permutation of elements of $\Omega$ by action of $G$. Then show the center of $G$ is not trivial.
I know that we can work for each permutation, there are multiple possible $h \in H$ since $h$ acts on a set. Also, we can count how many permutations of the set there are, but now I am stuck.
Label the elements of $\Omega$ as $I:=\{1,2\}$, $II:=\{3,4\}$, $III:=\{5,6\}$, $IV:=\{7,8\}$.
It is clear that $\Phi(gh)=\Phi(g)\Phi(h)$, so we have a homomorphism from $G$ into the symmetric group on the four letters $I,II,III,IV$.
But $\Phi$ is onto since (a) every transposition occurs in the image, for example $(I\ II)=\Phi((13)(24))$; (b) every permutation is a product of transpositions.
The kernel of $\Phi$ is easy to find; it is the set of permutations fixing each of the four sets $I,II, III, IV$ which is just $\langle (12), (34), (56), (78) \rangle$, an abelian group of order $16$.
Hence by the Isomorphism Theorem $|G|= 16. 4!$.
To see the centre is not trivial it is enough to find one non-trivial central element. Consider $(12)(34)(56)(78)$. For any $g\in G$ the element $g^{-1}(12)(34)(56)(78) g$ (a) lies in $\ker\Phi$, since kernels are normal subgroups and (b) is a product of $4$ transpositions, since conjugacy preserves cycle-shape. But $\ker\Phi=\langle (12), (34), (56), (78)\rangle$ has only one element which is a product of $4$ transpositions, so that $g^{-1}(12)(34)(56)(78) g=(12)(34)(56)(78)$, and so $(12)(34)(56)(78)$ is central in $G$.