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March 18th, 2018, 04:01 AM  #1 
Member Joined: Jan 2016 From: Blackpool Posts: 95 Thanks: 2  Order of a group question
Hello I am struggling to understand the following definition: If p is prime and $\mathbb{Z}_{p}$ then the group $(\mathbb{Z}_{p}*,\times)$ is a group of order p1. Why is this the case? I thought the order of a group mod p was just the number of elements in the group? Thanks. EDIT: I think it's because it's a group with respect to multiplication not including 0? Can someone confirm? Last edited by skipjack; March 18th, 2018 at 08:08 AM. 
March 18th, 2018, 08:10 AM  #2 
Global Moderator Joined: Dec 2006 Posts: 19,190 Thanks: 1649 
That would be my understanding.

March 18th, 2018, 11:26 AM  #3  
Senior Member Joined: Aug 2012 Posts: 1,960 Thanks: 547  Quote:
What's the order of, say, $(\mathbb Z_{10})^*$? What's the underlying principle that unifies these two examples? Last edited by Maschke; March 18th, 2018 at 11:30 AM.  
March 18th, 2018, 05:30 PM  #4  
Senior Member Joined: Sep 2016 From: USA Posts: 398 Thanks: 212 Math Focus: Dynamical systems, analytic function theory, numerics  Quote:
Last edited by skipjack; March 24th, 2018 at 03:56 AM.  
March 24th, 2018, 02:55 AM  #5 
Math Team Joined: Jan 2015 From: Alabama Posts: 3,248 Thanks: 887 
Take a simple example. If p= 3, the elements of $\displaystyle Z_3$ are 1 and 2. 0 is, of course, not in this group, as you said, since 0 has no multiplicative inverse.

March 24th, 2018, 04:06 AM  #6 
Global Moderator Joined: Dec 2006 Posts: 19,190 Thanks: 1649 
The usual notation is $(\mathbb{Z}_{p}^*,\times)$, which is possibly what Jaket1 intended. The elements of $\mathbb{Z}_3$ are 0, 1 and 2, whereas the elements of $\mathbb{Z}_3^*$ are 1 and 2. 
March 24th, 2018, 05:04 AM  #7  
Senior Member Joined: Aug 2017 From: United Kingdom Posts: 204 Thanks: 60 Math Focus: Algebraic Number Theory, Arithmetic Geometry  Quote:
If you're talking about $(\mathbb{Z}_{p}^*, \times)$, then (the class of) $p$ isn't even an element of this group  $p$ has no multiplicative inverse mod $p$. That this group has order $p1$ is a simple consequence of Bezout's identity. To see that it's cyclic is also quite elementary, but is a bit more involved  it takes more than just FLT.  
March 26th, 2018, 09:15 AM  #8 
Senior Member Joined: Mar 2015 From: New Jersey Posts: 1,392 Thanks: 100 
The OP is unambiguous (typo): $\displaystyle (\mathbb{Z}^{*}_{p},\times)$ Absence of 0 is the result of modular algebra for multiplication. https://www.dimgt.com.au/multiplica...oupmodp.html 
March 26th, 2018, 10:07 AM  #9  
Senior Member Joined: Aug 2017 From: United Kingdom Posts: 204 Thanks: 60 Math Focus: Algebraic Number Theory, Arithmetic Geometry  Quote:
 
March 27th, 2018, 06:05 AM  #10 
Senior Member Joined: Mar 2015 From: New Jersey Posts: 1,392 Thanks: 100 
You could also ask for the order of an element a of a group: smallest n st $\displaystyle a^{n}=1$. $\displaystyle \mathbb{Z}^{*}_{5}$ for example, $\displaystyle 3^{2}\equiv 4 (mod5), 3^{3}\equiv 12 (mod5) \equiv 2 (mod5), 3^{4} \equiv 6 (mod5) \equiv 1(mod5)$, so 3 has order 4, $\displaystyle 0\times a \equiv 1(mod5)$ has no solution (0 has no inverse) so 0 is not in the group, which consists of 1,2,3,4. If you do the multiplication table for 1,2,3,4, every element has an inverse. 

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