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Does every permutation of group generators imply an automorphism?

  1. Jul 31, 2013 #1
    I couldn't find the words to summarize my question perfectly in the title so I will clarify my question here.

    Say we have a group G in which every element can be written in the form [itex]g_1^{e_1} g_2^{e_2}...g_n^{e_n}, 0 ≤ e_i < |g_i| [/itex].

    Suppose that there exists a different set [itex] g_1', g_2', ..., g_n' [/itex] that generates G in the same way: [itex]g = g_1'^{e'_1} g_2'^{e'_2}...g_n'^{e'_n}, 0 ≤ e'_i ≤ |g'_i| [/itex] where [itex] |g'_i| = |g_i| [/itex]. (As a concrete example, think of Dn, in which every element can be written as [itex]s^i r^j, 0 ≤ i < 2, 0 ≤ j < n [/itex] and r can be any rotation of order n, and s can be any reflection).

    Then suppose [itex] \phi: G → G [/itex] is an automorphism. Then [itex] \phi [/itex] is completely determined by its effect on the generators of G. But are we guaranteed that switching [itex]g_i, g'_i[/itex] always guarantees an automorphism, and that all values of [itex]\phi [/itex] for the generators are independent?

    For example, in Dn, can I define [itex]\phi(r)[/itex] to be any rotation with order n, and define [itex]\phi(s)[/itex] to be any reflection? Does the choice of [itex]\phi(s)[/itex] depend on [itex]\phi(r)[/itex]?
     
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  3. Jul 31, 2013 #2

    micromass

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    Here's a nice counterexample. Consider the group ##S_4##, this can be visualized as:

    600px-Symmetric_group_4%3B_Cayley_graph_4%2C9.svg.png

    The blue arrow is just applying ##B= (1~2~3)##. The red arrows are applying ##R = (1~4~3~2)##.

    The entire group structure is visible from this diagram. For example, we see that ##B^3 = I##. Indeed, pick an arbitrary point (for example ##1234## in the bottom left corner), follow the blue arrows 3 times and you end up at the same point. In the same way, we see that ##R^4 = I##. Furthermore, we see easily that this generates the group.

    Now, consider ##\varphi(B) = RRBB## (I use the usual composition notation, thus: follow blue two times, then follows red two times). We see that ##(RRBB)^3 = I##.
    Furthermore, we send ##\varphi(R) = BRB##. We see that ##(BRB)^4 = I##.

    However, we can easily check that ##RBR = BB##. But
    [tex]\varphi(RBR) = BRBRRBBBRB \neq RRBBRRBB = \varphi(BB)[/tex]

    Indeed, if we start from ##1234## and we follow ##BRBRRBBBRB##, then we end up with ##1432##. And if we follow ##RRBBRRBB##, then we end up with ##4132##.

    All of this can be checked algebraically as well, but it's tedious.

    The difference with your example is that your example, we could write any example in a unique way in the form ##r^a s^b##. This decomposition is not unique in my ##S_4## example. The difference is essentially that the dihedral group is a semi-direct product, while ##S_4## is not.
     
  4. Jul 31, 2013 #3
    Right, and then [itex]\phi[/itex] isn't well-defined. Ok, so we need to require that the factorization be unique?
    About [itex]S_4[/itex], I have the conjecture that you can replace the generator with another generator of the same order reachable by at most one more "color-change" than it, starting from the identity. Like, for BB = (231) we can only switch it with a generator that has at most one color change. Like B B* R* where * means repeat an arbitrary number of times.

    EDIT: ah no nevermind that. I don't think it works.
     
    Last edited: Jul 31, 2013
  5. Jul 31, 2013 #4

    micromass

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    I'm not sure what your conjecture is really. But this should help: the automorphisms of ##S_4## are exactly of the form

    [tex]\varphi:S_4\rightarrow S_4:g\rightarrow hgh^{-1}[/tex]

    where ##h\in S_4## is fixed. Thus the possible elements that ##B## (or any other element) can be sent to by an automorphism is given by ##\{hBh^{-1}~\vert~h\in S_4\}##. This is the conjugacy class of ##B##: http://en.wikipedia.org/wiki/Conjugacy_class
     
  6. Jul 31, 2013 #5
    yeah what I said didn't really make sense.

    So for S4 the set of all automorphisms is the same as the setof inner automorphisms?

    If the factorization is unique, as it is for Dn, is it true that the elementary rotation can be sent to any other rotation of order n, and any reflection can be sent to any other reflection?
     
  7. Jul 31, 2013 #6

    micromass

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    Indeed.

    My gut feeling says yes, but you probably want to prove this.
     
  8. Jul 31, 2013 #7
    I'm not really sure. I actually think that we need to add the requirement that the two sets of generators behave in the same way when multiplied. For Dn, for example, we should also say that s'r's' = r'^(-1).

    In this case this is how I would try to prove it:

    Suppose [itex] g_1, ..., g_n [/itex] are the generators, and every element of G can be factored in a unique way in the form [itex] g = g_1^{e_1}...g_n^{e_n} [/itex]. Then there is a one-to-one correspondence between g and the n-tuple [itex] (e_1, ..., e_n ) [/itex] so I'll use the n-tuple as shorthand. Then we must look at how [itex] (e_1, ..., e_n ) [/itex] behaves when multiplied by another n-tuple. We can define the product of two n-tuples as being the n-tuple consisting of the exponents of the [itex]g_i[/itex] after multiplying the corresponding elements. For example in abelian groups, this is simply the addition operation in [itex]Z_{|g_1|} × ... × Z_{|g_n|} [/itex].
    For [itex]D_n[/itex], [itex]s^ir^j \rightarrow (i, j) [/itex] and [itex] (i, j) * ( i', j') = ( i + i' (mod 2), (-1)^{i'}j + j' (mod n) ) [/itex] corresponding to the fact that [itex] sr = r^{-1}s [/itex].

    In both cases [itex] (e_1, ... e_n ) * ( e'_1, ... e'_n ) = ( f_1( e_1, ..., e_n, e'_1, ..., e'_n ), ..., f_n( e_1, ... e_n, e'_1, ..., e'_n ) ) [/itex].

    So the requirement is that [itex] f_1, ..., f_n [/itex] be the same for both sets of generators.

    In that case, let [itex] \phi [/itex] be an automorphism which sends the generating set to another set with the same [itex] f_i [/itex]. Then by the previous equation, [itex]\phi( (e_1, ..., e_n ) * ( e'_1, ..., e'_n ) ) = \phi( ( f_1( e_1, ..., e_n, e'_1, ..., e'_n ), ..., f_n( e_1, ... e_n, e'_1, ..., e'_n ) ) [/itex] [itex] = ( f_1( e_1, ..., e_n, e'_1, ..., e'_n ), ..., f_n( e_1, ... e_n, e'_1, ..., e'_n ) ) = ( e_1, ..., e_n ) * ( e'_1, ..., e'_n ) = \phi( ( e_1, ..., e_n ) ) * \phi( ( e'_1, ..., e'_n ) ) [/itex].

    So it is indeed an automorphism.

    (To clarify a bit, 'cause I know this was very messy, the n-tuples represent only the exponents. The automorphism only changes the generators, not the exponents so this is why the n-tuples don't change after applying phi)
     
    Last edited: Jul 31, 2013
  9. Jul 31, 2013 #8
    So if for some reason, there were elements s' and r' of order 2 and n in Dn that didn't follow s'r' = r'^(-1)s then we would not be allowed to switch them with the generators s and r.

    We can check every pair of generators [itex] g_i, g_j [/itex] and see how their exponents change after commuting them: [itex] g_i * g_j = g_i^{x_{ij}}g_j^{x_{ji}} [/itex]. Then the functions are essentially determined by all values of [itex] x_{ij}, x_{ji} [/itex].
     
    Last edited: Jul 31, 2013
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