# Find the order of the multiplicative group of integers modulo 210

How do i find the order of the multiplicative group of integers modulo 210 ?

I want to use the order to prove that the multiplicative group of integers modulo 210 cannot have subgroups of order 6, 10 and 14 by Lagrange theorem.

Also, how do i prove the closure property for the multiplicative group of integers modulo 210 group.

• What do you mean by the closure property? Anyway, first you need to tell us which definition of the group you have been given. – Tobias Kildetoft Oct 24 '16 at 11:08
• Are you familiar with Euler's totient function? – Tobias Kildetoft Oct 24 '16 at 11:13
• Not really, i only know what it does. – Little Rookie Oct 24 '16 at 11:14
• Well, that is a good place to start (you could of course just count this manually if you really have never seen the properties of the totient fiunction, but that does not seem like the intended wa to do this). – Tobias Kildetoft Oct 24 '16 at 11:18
• @yh05 Note that $11$ has order $6$ modulo $210$. What leads you to believe it's possible to show that the group of integers modulo $210$ doesn't contain a subgroup of order $6$? – Erick Wong Oct 24 '16 at 15:33

$$\phi(210)=\phi(2\cdot3\cdot5\cdot7)=\phi(2)\phi(3)\phi(5)\phi(7)=1\cdot2\cdot4\cdot6=48$$
So $\;\left|\left(\Bbb Z/210\Bbb Z\right)^*\right|=48=2^3\cdot4\;$ . This is a general property of these gorups, by the way.
• Well, if you know that for $\;k\in\Bbb Z/210\Bbb Z\;$, then $\;k\in\left(\Bbb Z/210\Bbb Z\right)^*\iff (k,210)=1\;$ , then you can prove that $\;(k,210)=1\,,\,(m,210)=1\;\implies (km,210)=1\;$ – DonAntonio Oct 24 '16 at 11:28
• @yh05 No way can $\;km=210\;$ as both $\;k,\,m\;$ are coprime with $\;210\;$ ...! – DonAntonio Oct 24 '16 at 11:58