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Thread: Automorphism

  1. #1
    Jul 2011


    Show that Aut(Z/24Z) is isomorphic to Z/2Z (direct sum) Z/2Z (direct sum) Z/2Z (direct sum) Z/2Z. Not sure how to approach this problem. Help, please!
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  2. #2
    MHF Contributor
    Dec 2012
    Athens, OH, USA

    Re: Automorphism

    It depends on what you know about the automorphism group of a cyclic group.
    1. Aut($\displaystyle Z_n$) is isomorphic to the group of (multiplicative) units of $\displaystyle Z_n, Z_n^*$.

    2. If $\displaystyle n=p_1^{n_1}p_2^{n_2}...p_k^{n_k}$ is the factorization of n into primes, Aut($\displaystyle Z_n$) is isomorphic to $\displaystyle Z_{p_1^{n_1}}^*\oplus Z_{p_2^{n_2}}^*\oplus...Z_{p_k^{n_k}}^*$

    3. If p is odd $\displaystyle Z_{p^n}^*$ is isomorphic to $\displaystyle Z_{{p^{n-1}(p-1)}$

    4. A similar statement for n a power of 2 (I'm tired of typing Latex).

    All of the above is easily found on the web -- use Google

    I'll assume just 1. So we need the group multiplicative units of $\displaystyle Z_{24}$; i.e. the set of integers < 24 which are relatively prime to 24, namely {1,5,7,11,13,17,19,23}. Now
    $\displaystyle 1^2=1, 5^2=25=1, 7^2=49=1, 11^2=121=1, 13^2=(-11)^2=1$, etc (equality here is read mod 24). That is all elements have order 2. Now you must figure that out that such a group is the direct sum of 3 cyclic groups of order 2. One "high powered" result: If an abelian group G has order p^n for a prime p and every element has order p, G is isomorphic to the direct sum of n cyclic groups of order p. This is one part of the fundamental theorem of finite abelian groups.

    I really think the general proofs of the statements above are no harder than trying to do the specific n = 24.
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  3. #3
    Senior Member jakncoke's Avatar
    May 2010

    Re: Automorphism

    I'm not really 100% sure about my solution so take it with a grain of salt.

    So i hope you will accept when i say $\displaystyle \mathbb{Z}/24\mathbb{Z} \cong Z_{24} $.

    Now let us examine the structure of Automorphism group of the cyclic group, denoted $\displaystyle Aut(Z_{24}) $.

    I assert that every automorphism in Aut$\displaystyle \mathbb{Z}_{24}$) is determined by where it sends $\displaystyle 1 \in Z_{24} $. (or $\displaystyle \mathbb{Z}_{n} $ generally)
    For if $\displaystyle \phi_{i} \in Aut(\mathbb{Z}_{24}$ and $\displaystyle g \in G $ $\displaystyle \phi_{i}(g) = \phi_{i}(1^{g}) = \phi_{i}(1)^{g} $ (Also note that we showed every element can be expressed as a power of $\displaystyle \phi_{i}(1)$ and so $\displaystyle \phi_{i}(1)$ is a generator for G

    I also assert that if $\displaystyle \phi_{i}, \phi{j} \in Aut(\mathbb{Z}_{24}) $
    Then if $\displaystyle \phi_{i}(1) = m = \phi_{j}(1) $ then $\displaystyle \phi_{i} = \phi{j}$
    Now take $\displaystyle g \in G $ so $\displaystyle \phi_{i}(g) = \phi_{i}(1^g) = \phi_{i}(1)^{g} = m^{g} = mg $
    Similarly, $\displaystyle \phi_{j}(g) = \phi_{j}(1^g) = \phi_{j}(1)^{g} = m^{g} = mg $ so $\displaystyle \phi_{j} = \phi_{i} $

    Since we know $\displaystyle \phi_{i}(1)$ maps to generators of $\displaystyle \mathbb{Z}_{24} $ or elements of $\displaystyle U(24)$ which number $\displaystyle \phi(24) = 8 $
    Now there are only 3 groups of size 8. $\displaystyle \mathbb{Z}_{8} $ , $\displaystyle \mathbb{Z}_{2} \oplus \mathbb{Z}_{4} $ and $\displaystyle \mathbb{Z}_{2} \oplus \mathbb{Z}_{2} \oplus \mathbb{Z}_{2} $
    Now $\displaystyle \mathbb{Z}_{8} $ is cyclic, but $\displaystyle Aut(Z_{24}) $ isn't. geez. im tired. Im sorry to leave without finishing, hopefully this is somewhat correct. you can probably verify the last statements
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  4. #4
    MHF Contributor

    Mar 2011

    Re: Automorphism

    Z/24Z is cyclic. so any homomorphism (and automorphisms ARE homomorphism) from Z/24Z to itself is completely determined by the image of 1+24Z, which generates Z/24Z.

    since an automorphism must be bijective, the image of 1+24Z must likewise be a generator of Z/24Z. there are φ(24) = φ(3)φ(8) = 2*4 = 8 generators of Z/24Z. these are, specifically:

    1+24Z, 5+24Z, 7+24Z, 11+24Z, 13+24Z, 17+24Z, 19+24Z, 23+24Z.

    it should be clear (and it is easy to verify) that the map Z/24Z-->Z/24Z given by 1+24Z-->k+24Z (so that n+24Z-->kn+24Z) for k = 1,5,7,11,13,17,19 or 23 gives an automorphism, and these are the only ones possible.

    note that these automorphisms commute:

    if we denote n+24Z -->kn+24Z by φk, we have:

    φk∘φk'(n+24Z) = φkk'(n+24Z)) = φk(k'n+24Z) = k(k'n)+24Z = (kk')n+24Z = (k'k)n+24Z = k'(kn)+24Z = φk'(kn+24Z) = φk'k(n+24Z)) = φk'∘φk(n+24Z).

    so we have an abelian group of order 8. by the structure theorem for finite abelian groups this must be either:

    Z8, which is cyclic
    Z4xZ2, which has no elements of order 8, but does has elements of order 4
    Z2xZ2xZ2, which has only elements of order 2, and the identity.

    so to show that Aut(Z/24Z) is isomorphic to the last one, it suffices to show that every element of Aut(Z/24Z) has order 2, which is equivalent to showing that every non-identity unit of the ring Z/24Z has multiplicative order 2. and:

    5*5 = 25 = 1 (mod 24)
    7*7 = 49 = 1 (mod 24)
    11*11 = 121 = 1 (mod 24)
    13*13 = 169 = 1 (mod 24)
    17*17 = 289 = 1 (mod 24)
    19*19 = 361 = 1 (mod 24)
    23*23 = 529 = 1 (mod 24).


    the fast way:

    Aut(Z/nZ) ≅ U(n). when n = 24:

    U(24) = U(3) x U(8) = Z2 x (Z2 x Z2)

    (this requires knowing that U(8) = {1,3,5,7} under multiplication mod 8 is isomorphic to the klein 4-group (Z2 x Z2), which can be shown by showing both 3 and 7 are of order 2, since a cyclic group of order 4 must have two elements of order 4).
    Last edited by Deveno; Feb 8th 2013 at 05:55 PM.
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