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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- Feb 8th 2013, 02:22 PMlovesmathAutomorphism
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!

- Feb 8th 2013, 06:34 PMjohngRe: Automorphism
It depends on what you know about the automorphism group of a cyclic group.

1. Aut( ) is isomorphic to the group of (multiplicative) units of .

2. If is the factorization of n into primes, Aut( ) is isomorphic to

3. If p is odd is isomorphic to

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 ; i.e. the set of integers < 24 which are relatively prime to 24, namely {1,5,7,11,13,17,19,23}. Now

, 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. - Feb 8th 2013, 06:37 PMjakncokeRe: 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 .

Now let us examine the structure of Automorphism group of the cyclic group, denoted .

I assert that every automorphism in Aut ) is determined by where it sends . (or generally)

For if and (Also note that we showed every element can be expressed as a power of and so is a generator for G

I also assert that if

Then if then

Now take so

Similarly, so

Since we know maps to generators of or elements of which number

Now there are only 3 groups of size 8. , and

Now is cyclic, but isn't. geez. im tired. Im sorry to leave without finishing, hopefully this is somewhat correct. you can probably verify the last statements - Feb 8th 2013, 06:53 PMDevenoRe: 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 φ_{k}, we have:

φ_{k}∘φ_{k'}(n+24Z) = φ_{k}(φ_{k'}(n+24Z)) = φ_{k}(k'n+24Z) = k(k'n)+24Z = (kk')n+24Z = (k'k)n+24Z = k'(kn)+24Z = φ_{k'}(kn+24Z) = φ_{k'}(φ_{k}(n+24Z)) = φ_{k'}∘φ_{k}(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).