# Using Cayley's Theorem to find an isomorphic subgroup.

Find a subgroup of $S_4$ isomorphic to $\mathbb{Z} / 2\mathbb{Z} \times \mathbb{Z} / 2\mathbb{Z}$.

Here is what I have so far: We need to find a subgroup of $S_4$ that is isomorphic to $\{0, 1 \} \times \{0, 1 \}$. Now, $\{0, 1 \} \times \{0, 1 \} = \{(0,0), (0,1), (1,0), (1,1) \}$. The pairs $(0,0)$ and $(1,1)$ tell us that the identity map, $e$, is present. The pairs $(0,1)$ and $(1,0)$ tell us that the transposition map, $\tau$ is present. So a subgroup of $S_4$ isomorphic to $\mathbb{Z} / 2\mathbb{Z} \times \mathbb{Z} / 2\mathbb{Z}$ must be $\{e, \tau \}$.

Comment: This seemed too easy and I'm thinking there must be more. Thank you for your help!

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"the" transposition map? How many elements does $\{e,\tau\}$ have? How many does ${\bf Z}/2{\bf Z}\times{\bf Z}/2{\bf Z}$ have? –  anon Mar 19 '13 at 4:05
Wikipedia article on Klein four-group (a.k.a. Viergruppe) mentions permuation representation of this group. –  Martin Sleziak Mar 19 '13 at 7:57
Do you know GAP? If you like, I can give you the codes. ;-) –  Babak S. Mar 19 '13 at 17:17

The number $4$ is just large enough to pick two disjoint transpositions. Say $$\sigma=(1,2)\qquad\tau=(3,4).$$ Now check that the subgroup generated by $\sigma$ and $\tau$ does what you want. More generally, you see why $S_{2n}$ contains a subgroup isomorphic to $\mathbb{Z}_2^n$.

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If you want to use Cayley (or have to, by assignment), take the regular representation as follows. (I am really going through the proof of Cayley's theorem.)

Denote the elements $(0,0), (0,1), (1,0), (1,1)$ by $1, 2, 3, 4$ in order. Now look at the action of each of them by addition on $(0,0), (0,1), (1,0), (1,1)$.

Clearly $(0,0)$ acts as the identity $e$.

Now consider $(0,1)$. We have $$(0,0)+ (0,1) =(0,1),\quad (0,1)+ (0,1) =(0,0),\quad (1,0)+ (0,1) =(1,1),\quad (1,1)+ (0,1) =(1,0),$$ so this is the permutation $(12)(34)$, as $(0,0) \leftrightarrow (0,1)$ and $(1,0) \leftrightarrow (1,1)$.

Now consider $(1,0)$. We have $$(0,0)+ (1,0) =(1,0),\quad (0,1)+ (1,0) =(1,1),\quad (1,0)+ (1,0) =(0,0),\quad (1,1)+ (1,0) =(0,1),$$ so this is the permutation $(13)(24)$, as $(0,0) \leftrightarrow (1,0)$ and $(0,1) \leftrightarrow (1,1)$.

Now consider $(1,1)$. We have $$(0,0)+ (1,1) =(1,1),\quad (0,1)+ (1,1) =(1,0),\quad (1,0)+ (1,1) =(0,1),\quad (1,1)+ (1,1) =(0,0),$$ so this is the permutation $(14)(23)$, as $(0,0) \leftrightarrow (1,1)$ and $(0,1) \leftrightarrow (1,0)$.

So the subgroup you obtain is $$\{ e, (12)(34), (13)(24), (14)(23)\}.$$ Note that it is a regular subgroup of $S_4$, as always when taking the regular representation.

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Why dont you try another apporach.

In $\mathbb{Z}_2\times\mathbb{Z}_2$, all elements are of order $2$. In $S_4$, there are only $9$ elements are of order $2$. It is not hard to try some of the elements and get a subgroup that is isomorphic to $\mathbb{Z}_2\times\mathbb{Z}_2$.

One example could be $\{e, (12), (34), (12)(34)\}$ and there are a lot more.

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