Find the $9$ subgroups of $S_4$ such that no group on your list is isomorphic.

Question

I know that for a group to be Isomorphic It has to have the same order. So I started by trying to find a subgroup of each possible order (that divides the size of the original group) but I realized that it would take me forever if I tried to do it numerically. But I cant see any other way to do it.

Answer 1

The best way is to start listing generators until you run out of groups.

Trivial/improper subgroups:

• $\{0\}$

• $S_4$.

Subgroups generated by one element:

• $\langle (12)\rangle\simeq\mathbb{Z}/2$ (all transpositions generate this group)

• $\langle (123)\rangle\simeq\mathbb{Z}/3$ (all triples generate this group)

• $\langle (1234)\rangle\simeq\mathbb{Z}/4$

Subgroups generated by two elements:

• $\langle(12),(34)\rangle\simeq(\mathbb{Z}/2)^2$ (all disjoint pairs of transpositions generate this group).

• $\langle (12),(123)\rangle\simeq S_3$ (its support is on the set $\{1,2,3\}$, so this group is a subgroup of $S_3$).

• $\langle (13),(1234)\rangle\simeq D_4$. I find this one the hardest to spot, but the two elements listed above have the relations for $D_4$.

Special subgroups (some generated by two elements):

• $A_4$

Note also that $\langle (12)(34),(13)(24)\rangle$ form a normal subgroup isomorphic to $(\mathbb{Z}/2)^2$.