Find two non-isomorphic subgroups of $S_4$ that have order 4.

Question

Find two subgroups of $S_4$ that have order 4, such that they are not isomorphic to each other.

Is there a convenient way to obtain the answer instead of writing out the permutations of $S_4$? How can two subgroups of $S_4$ be non-isomorphic to each other? (Since both of them are abelian... Perhaps one is cyclic and the other not?)

Answer 1

The approach is right. For the non-cyclic subgroup, try the group generated by $(12)$ and $(34)$.

It should be easy to produce a cyclic subgroup of order $4$.

Answer 2

Consider $G_1=<(1 \ 2 \ 3 \ 4)>, G_2=<(1 \ 2),(3 \ 4)>$

$G_1$, $G_2$ are both of order 4 and are both subgroups of $S_4$, but $G_1$ is cyclic and $G_2$ isn't, hence they are non-isomorphic

Answer 3

For the future, you should work on growing a catalog of groups with which you're familiar. The two that come in handy here are

• The cyclic group of order $4$. If you can find an element $g$ of order $4$ in any group, then the subgroup $\langle g \rangle$ generated by $g$ is cyclic of order $4$. The other group that we'll care about is

• The Klein Four Group. This group has order $4$ but is not cyclic. You can find it if you can find two elements $x$ and $y$ of order $2$ that commute with each other. Then the subgroup $\langle x, y \rangle$ generated by $x$ and $y$ will be isomorphic to the Klein Four Group (a direct product of cyclic groups of order $2$) which is not isomorphic to the cyclic group of order $4$.

I know the other answers have mentioned essentially this, but my point is to maintain a list of groups that you learn well - cyclic groups, products of cyclic groups, dihedral groups, symmetric groups, etc. This list will come in handy, as you study group theory.

Answer 4

For me the key point here is Cayley's theorem which tells us that every four element group is isomorphic to a subgroup of the symmetric group on four elements. Furthermore the proof tell us how to find which subgroups.

The two four element groups are $\mathbb{Z}_4$ and $\mathbb{Z}_2\times\mathbb{Z}_2$ choosing bijections $\phi$ and $\psi$ between their underlying sets and the set $\{1,2,3,4\}$ defined by $\phi([0])=1,\phi([1])=2, \phi([2])=3,\phi([3])=4$ and $\psi(([0],[0])=1,\psi(([0],[1]))=2, \phi(([1],[0]))=3,\phi(([1],[1]))=4$ we obtain the two desired subgroups $$S_1=\{f_x\,|\,x \in \mathbb{Z}_4\}$$ $$S_1=\{g_y\,|\,y \in \mathbb{Z}_2\}$$ where $f_x$ and $g_y$ are defined for each $i$ in $\{1,2,3,4\}$ by $f_x(i)=\phi(x\phi^{-1}(i))$ and $g_y(i)=\psi(y\psi^{-1}(i))$ respectively.