I have to find the number of elements in $\operatorname{Aut}(\mathbb Z_{720})$ with order $6$. Please suggest how to go about it.

1) $\operatorname{Aut}(\mathbb Z_{720})$ 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 $\mathbb Z_n$.

In this case, $720 = 16 \times 9 \times 5$.

Therefore, $\operatorname{Aut}(\mathbb Z_{720})\cong U(720) \cong \mathbb Z_2 \times \mathbb Z_4 \times \mathbb Z_4 \times \mathbb Z_6$.

Now, the possible orders of elements in $\mathbb Z_2: 1, 2$; $\mathbb Z_4: 1,2,4$; $\mathbb Z_6: 1, 2, 3, 6$.

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

If $6= \operatorname{Order}(a, b, c, d) = \operatorname{lcm} ( \operatorname{Order} (a), \operatorname{Order} (b), \operatorname{Order} (c), \operatorname{Order} (d))$ then:

Case 1 : If $\operatorname{Order} (d) =6$ then $\operatorname{lcm} ( \operatorname{Order}(a), \operatorname{Order}(b), \operatorname{Order}(c))= 1$ or $2$.

Using the the result for cyclic groups:

for every divisor $d$ of the order of a cyclic group G, there exists $\phi(d)$ 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.

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

According to this, in each case, there can be 14 elements in all. Total no. of elements 16 + 14 = 30.

  • 3
    $\begingroup$ Your general approach looks good to me. But what about Case 2? $\endgroup$ – Jyrki Lahtonen Aug 26 '11 at 12:32

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