If p is prime, prove the group (Zp*,x) has exactly one element of order 2.

[ae1709]

Hi. I need to: prove the group (Zp*,x) has exactly one element of order 2. Here, p is prime and (Zp*,x) is the set {1, 2,..., p-1} under multiplication modulo p. Any help would be much appreciated!

 

[Office_Shredder]

I think the easier half is: which element has order two?

It might help if you write down the condition of what it means to have order 2

 

[micromass]

Something that might help you is knowing that $\mathbb{Z}_p^*$ is cyclic. Do you know this already??

 

[micromass]

If you do not know the previous. Then you might want to think about the polynomial $X^2-1$ in $\mathbb{Z}_p[X]$. What do you know about its roots??

 

[blue_raver22]

Sure, I would be happy to provide a response to this content.

To prove that the group (Zp*,x) has exactly one element of order 2, we first need to understand what it means for an element to have order 2 in a group. The order of an element in a group is the smallest positive integer n such that the element raised to the power of n equals the identity element (usually denoted as e). In other words, the order of an element is the number of times it needs to be multiplied by itself to get the identity element.

Now, in the group (Zp*,x), the identity element is 1, and since p is prime, all the elements in this group are relatively prime to p. This means that for any element a in the group, a raised to the power of p-1 (which is the number of elements in the group) will always equal 1. This is known as Fermat's Little Theorem.

So, if an element a has order 2 in this group, it means that a^2 = 1, or in other words, a is its own inverse. This also means that a = a^-1. Now, since a is relatively prime to p, we can use the Extended Euclidean algorithm to find its inverse. This algorithm will give us a unique inverse for each element in the group. Therefore, if an element has order 2, its inverse will also have order 2.

Now, we know that the identity element 1 has order 1, and we have just established that any element with order 2 will have an inverse with order 2. This means that if there is more than one element with order 2 in the group, there will be more than two elements with order 1 (since the inverse of an element with order 2 will also have order 2). But this is not possible, since the only element with order 1 in this group is the identity element.

Therefore, we can conclude that there can only be one element with order 2 in the group (Zp*,x). This completes the proof. I hope this helps! Let me know if you have any further questions.