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 March 18th, 2018, 05:01 AM #1 Senior Member   Joined: Jan 2016 From: Blackpool Posts: 100 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 p-1. 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 09:08 AM.
 March 18th, 2018, 09:10 AM #2 Global Moderator   Joined: Dec 2006 Posts: 19,984 Thanks: 1853 That would be my understanding. Thanks from Jaket1
March 18th, 2018, 12:26 PM   #3
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Quote:
 Originally Posted by Jaket1 EDIT: I think it's because it's a group with respect to multiplication not including 0? Can someone confirm?
True, but why? It's not a definition, it needs proof. How do you know the nonzero elements form a multiplicative group?

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 12:30 PM.

March 18th, 2018, 06:30 PM   #4
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Quote:
 Originally Posted by Jaket1 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 p-1. 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?
The number of elements in any group is its order. That is the definition of order. As for why $Z_p$ has order $(p-1)$, this is just Fermat's little theorem, which guarantees that $p$ generates a cyclic subgroup of $Z_p$ of order $(p-1)$.

Last edited by skipjack; March 24th, 2018 at 04:56 AM.

 March 24th, 2018, 03:55 AM #5 Math Team   Joined: Jan 2015 From: Alabama Posts: 3,261 Thanks: 894 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, 05:06 AM #6 Global Moderator   Joined: Dec 2006 Posts: 19,984 Thanks: 1853 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, 06:04 AM   #7
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Quote:
 Originally Posted by SDK The number of elements in any group is its order. That is the definition of order. As for why $Z_p$ has order $(p-1)$, this is just Fermat's little theorem, which guarantees that $p$ generates a cyclic subgroup of $Z_p$ of order $(p-1)$.
Would you mind clarifying what you mean here? If you're talking about $(\mathbb{Z}_{p}, +)$, then this is a cyclic group of order $p$ generated by the class of $1$. The class of $p$ is the identity element, so generates the trivial subgroup.

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 $p-1$ 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, 10:15 AM #8 Senior Member   Joined: Mar 2015 From: New Jersey Posts: 1,603 Thanks: 115 The OP is unambiguous (typo): $\displaystyle (\mathbb{Z}^{*}_{p},\times)$ Absence of 0 is the result of modular algebra for multiplication. https://www.di-mgt.com.au/multiplica...oup-mod-p.html
March 26th, 2018, 11:07 AM   #9
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 Originally Posted by zylo The OP is unambiguous (typo): $\displaystyle (\mathbb{Z}^{*}_{p},\times)$ Absence of 0 is the result of modular algebra for multiplication. https://www.di-mgt.com.au/multiplica...oup-mod-p.html
I was replying to SDK, not the OP.

 March 27th, 2018, 07:05 AM #10 Senior Member   Joined: Mar 2015 From: New Jersey Posts: 1,603 Thanks: 115 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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