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Cayley's Theorem (isomorphisms of Cyclics to SN)

  1. Feb 9, 2015 #1


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

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

    Is it correct to say then that [itex]C_3 \cong A_3 [/itex] ? the orders seem to match.

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

    1. [itex] \{ e \}[/itex]
    2. [itex] S_4 [/itex]
    3. [itex] \{ (.) \} \cup \{ (..)(..) \} \cong C_2 \times C_2 [/itex]
    4. [itex] \{ (.) \} \cup \{ (...) \} \cup \{ (..)(..) \} \cong A_4[/itex]

    My problem is that in this case, the only "order" which matches is the [itex]C_2 \times C_2[/itex], 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. jcsd
  3. Feb 9, 2015 #2


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    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.
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