Cayley's Theorem (isomorphisms of Cyclics to SN)

[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?

 

[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.