# 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. Commented Oct 24, 2016 at 11:08
• Are you familiar with Euler's totient function? Commented Oct 24, 2016 at 11:13
• Not really, i only know what it does. Commented Oct 24, 2016 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). Commented Oct 24, 2016 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 $6$? Commented Oct 24, 2016 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 $\;(k,210)=1\,,\,(m,210)=1\;\implies (km,210)=1\;$ Commented Oct 24, 2016 at 11:28
• @yh05 No way can $\;km=210\;$ as both $\;k,\,m\;$ are coprime with $\;210\;$ ...! Commented Oct 24, 2016 at 11:58