Finding Elements of Order 6 in Aut(Z720)

[mehtamonica]

I have to find the number of elements in Aut(Z720) with order 6. Please suggest how to go about it.

1) Aut(Z720) isomorphic to U(720) (multiplicative group of units).

2 ) I am using the fundamental theorem of abelian group that a finite abelian group is isomorphic to the direct products of cyclic groups Zn.

In this case, 720=16×9×5.

Therefore, Aut(Z720)≅U(720)≅Z2×Z4×Z4×Z6.

Now, the possible orders of elements in Z2:1,2; Z4:1,2,4; Z6:1,2,3,6.

Using the result defining the order of an element in external direct products:

If 6=Order(a,b,c,d)=lcm(Order(a),Order(b),Order(c),Order(d)) then:

Case 1 : If Order(d)=6 then lcm(Order(a),Order(b),Order(c))=1 or 2.

Using the the result for cyclic groups:

for every divisor d of the order of a cyclic group G, there exists ϕ(d) elements in G with order d.

It seems there are 16 elements. I am not sure though.

Is this the correct way and how to proceed further? Please suggest.

 

[micromass]

Seems correct til now.

Now, another case that can pop up is when d has order 3. How many possibilities are there for d to have order 3? And what can a,b and c be then?

 

[mehtamonica]

Seems correct til now.

Now, another case that can pop up is when d has order 3. How many possibilities are there for d to have order 3? And what can a,b and c be then?

Case 2: If order of d = 3, then we need that lcm ( O(a), O(b), O(c))= 2

It can happen in three ways:

(a) O(a) = 2, O(b) = 1 or 2, O(c)= 1 or 2. (b) O(b) =2, O(a) = 1 or 2, O(c)= 1 or 2. (c) O(c)=2, O(a) = 1 or 2, O(b)= 1 or 2.

According to this, in each case, there can be 8 elements in all.

Total no. of elements 16 + 24 = 40.

Please suggest if it is correct .

 

[micromass]

Case 2: If order of d = 3, then we need that lcm ( O(a), O(b), O(c))= 2

It can happen in three ways:

(a) O(a) = 2, O(b) = 1 or 2, O(c)= 1 or 2. (b) O(b) =2, O(a) = 1 or 2, O(c)= 1 or 2. (c) O(c)=2, O(a) = 1 or 2, O(b)= 1 or 2.

According to this, in each case, there can be 8 elements in all.

Total no. of elements 16 + 24 = 40.

Please suggest if it is correct .

No, firstly there are multiple elements of order 3 in \mathbb{Z}_6. Furthermore, you have counted some elements multiple times.

For example a=b=c=2 occurs in cases (a), (b) and (c). So you have counted that three times.

 

[mehtamonica]

No, firstly there are multiple elements of order 3 in \mathbb{Z}_6. Furthermore, you have counted some elements multiple times.

For example a=b=c=2 occurs in cases (a), (b) and (c). So you have counted that three times.

ase 2: If order of d = 3, then we need that lcm ( O(a), O(b), O(c))= 2

It can happen in three ways:

(a) O(a) = 2, O(b) = 1 or 2, O(c)= 1 or 2. (b) O(b) =2, O(a) = 1 , O(c)= 1 or 2. (c) O(c)=2, O(a) = 1 , O(b)= 1.

According to this,

Total no. of elements 16 + 8 + 4 +2 = 30.

Please suggest if it is correct now.

 

[micromass]

ase 2: If order of d = 3, then we need that lcm ( O(a), O(b), O(c))= 2

It can happen in three ways:

(a) O(a) = 2, O(b) = 1 or 2, O(c)= 1 or 2. (b) O(b) =2, O(a) = 1 , O(c)= 1 or 2. (c) O(c)=2, O(a) = 1 , O(b)= 1.

According to this,

Total no. of elements 16 + 8 + 4 +2 = 30.

Please suggest if it is correct now.

That seems to be correct now!

 

[mehtamonica]

That seems to be correct now!

Thanks, you are a great teacher.

 

[lpetrich]

Here's how I'd solve that problem.

Z(720) is cyclic, meaning that it can be generated with one of its elements, a. That element has the property that a^k = identity only if k is a multiple of 720.

It can also be generated by every a^k, where k is relatively prime to 720.

So its automorphism group consists of all mappings a -> a^k, where k is relatively prime to 720. There are 192 possible k's, meaning that Aut(Z(720)) has 192 elements.


To go further, look for prime factors of 720: 2⁴ * 3² * 5. Thus,

Z(720) ~ Z(16) * Z(9) * Z(5)

Every element of Z(720) can be turned into a triplet of elements, one of each of the three groups on the right. A generator of Z(720) is thus
(generator of Z(16), generator of Z(9), generator of Z(5))

The same arguments about being relatively prime apply here also, and Aut(Z(16)) has order 8, Aut(Z(9)) has order 6, and Aut(Z(5)) has order 4.
Aut(Z(16)) ~ Z(4) * Z(2)
Aut(Z(9)) ~ Z(2) * Z(3)
Aut(Z(5)) ~ Z(4)
Aut(Z(720)) ~ Z(2)² * Z(4)² * Z(3)

We want to count how many elements of Aut(Z(720)) have order 6. That order is a product of 2 and 3, meaning that those elements can be found in this subgroup of Aut(Z(720)):
Z(2)⁴ * Z(3)
Doing careful counting yields (2⁴ - 1) * (3 - 1) = 15 * 2 = 30