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 ${\rm Aut}(\Bbb Z_{70}).$

  • 4
    $\begingroup$ What's $U(70)$ ? $\endgroup$ Feb 4 at 19:03
  • 2
    $\begingroup$ 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. $\endgroup$
    – azif00
    Feb 4 at 19:12
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    $\begingroup$ It's the group of units modulo $70$, @SassatelliGiulio. $\endgroup$
    – Shaun
    Feb 4 at 19:48
  • $\begingroup$ @azif00 youre right thanks $\endgroup$
    – eva33
    Feb 4 at 19:49

1 Answer 1


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.

  • $\begingroup$ i am not sure i understand how this proves that the group is abelian though $\endgroup$
    – eva33
    Feb 4 at 19:47
  • $\begingroup$ 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. $\endgroup$
    – Arthur
    Feb 4 at 19:52

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