Cayley's Theorem (isomorphisms of Cyclics to SN)

1. Feb 9, 2015

ChrisVer

Any finite group $G$ with order $|G| = N < \infty$ is isomorphic ($\cong$ ) to a subgroup of the symmetric group $S_N$.

I have one question here, let's take the cyclic group of order three: $C_3 = \{ e, c, c^2 \}$. By Cayley's theorem, this should be isomorphic to some subgroup of $S_3$.
The subgroups of $S_3$ are pretty definite, and they are also normal subgroups , created by the union of conjugacy classes. They are the $\{ (.) \}$ , $S_3$ (both trivial) and the union of $\{ (.) \} \cup \{ (...) \}$ which is the Alternating group $A_3$.

Is it correct to say then that $C_3 \cong A_3$ ? the orders seem to match.

On the other hand what happens if I go for example to $C_4$ ? In that case the subgroups of $S_4$ will be:

1. $\{ e \}$
2. $S_4$
3. $\{ (.) \} \cup \{ (..)(..) \} \cong C_2 \times C_2$
4. $\{ (.) \} \cup \{ (...) \} \cup \{ (..)(..) \} \cong A_4$

My problem is that in this case, the only "order" which matches is the $C_2 \times C_2$, But I don't think that the Klein 4-group is isomorphic to the cyclic group, because the orders of the elements don't match. Any help?

2. Feb 9, 2015

mathwonk

are you familiar with cycle notation? isn't (1234) a cycle of order 4, generating a cyclic subgroup of S4 of order 4? e.g. take the subgroup of permutations of the 4 vertices of a square, generated by the rotations of the square about its center, through multiples of 90 degrees.

the problem seems to be with your claim: " In that case the subgroups of S4 will be:...."

And you probably meant to post this in the abstract algebra section.