# Showing ${\rm Aut}(\Bbb Z_{70})$ is abelian

I am trying to prove that $${\rm Aut}(\Bbb Z_{70})$$ is abelian so I am thinking of proving that it is cyclic since cyclic groups are abelian.

We know that $${\rm Aut}(\Bbb Z_{70})$$ is isomorphic to $$U({70})$$ but also that $$U(m)$$ is cyclic only if $$m=2,4,p,2p^r$$, where $$p$$ is prime. $$70$$ cannot be written in this form so it is not cyclic right?

I'm kind of stuck here, any ideas?

Another thought of mine was to show that the order of $${\rm Aut}(\Bbb Z_{70})$$ is a prime number which means that the group is cyclic but i dont know how to calculate the order of $${\rm Aut}(\Bbb Z_{70}).$$

• What's $U(70)$ ? Feb 4 at 19:03
• Every cyclic group is abelian, but there are abelian groups which are not cyclic. If this is one of them, then what you’re trying to prove is impossible. Feb 4 at 19:12
• It's the group of units modulo $70$, @SassatelliGiulio. Feb 4 at 19:48
• @azif00 youre right thanks Feb 4 at 19:49

The automorphism group turns out to be $$\Bbb Z_4\times \Bbb Z_6$$, which is not cyclic, so your approach will not work, but it is clearly abelian. Calculating this group is, I think, most easily done with the Chinese remainder theorem, which tells you that $$\Bbb Z_{70}\cong \Bbb Z_2\times\Bbb Z_5\times \Bbb Z_7$$ (either as a group with only addition, or as a ring with addition and multiplication, or as a monoid with only multiplication).
Using that an element of a product ring or a product monoid is invertible iff each component is invertible, this approach ultimately gives you that the automorphism group is isomorphic to $$U_2\times U_5\times U_7$$, which is isomorphic to $$\Bbb Z_4\times \Bbb Z_6$$.
But you don't need to know exactly what the automorphism group is, how large it is, or what order its elements have. You say that know that $$\operatorname{Aut}(\Bbb Z_{70})$$ is isomorphic to $$U_{70}$$, the multiplicative group consisting of the invertible elements of $$\Bbb Z_{70}$$. The operation of said group is the multiplication of $$\Bbb Z_{70}$$, restricted to the relevant subset. Presumably you know that this multiplication is commutative on the whole of $$\Bbb Z_{70}$$. This almost immediately implies it is commutative on a subset.
• You mean you don't understand how to prove that $\Bbb Z_4\times \Bbb Z_6$ is abelian? You mean you don't understand how a commutative binary relation, when restricted to a subset, is still commutative? Then this exercise is not for you yet. Feb 4 at 19:52