Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Let $\mathbb{Z}_n ^*$ be the set of all units in $\mathbb{Z}_n$ and $(\mathbb{Z}_n ^*)^2$ = $\{ a^2 | a \in \mathbb{Z}_n ^*\} $

Then, decompose the factor group $\mathbb{Z}_n ^* / (\mathbb{Z}_n ^*)^2$

1.when $n$ is an odd prime

2.when $n$ is a product of two distinct odd primes $p, q$

I solved the first as following: In $\mathbb{Z}_p$, since $x^2=(p-x)^2$ holds, the order of $(\mathbb{Z}_n ^*)^2$ is exactly the half of that of $\mathbb{Z}_n ^*$.

Thus, the answer is $\mathbb{Z}_2$.

However the second is more complicated. I didn't know that. Please help me to solve this.

share|improve this question

2 Answers 2

Hints:

With your notation:

$$\Bbb Z_{pq}^*=\Bbb Z^*_p\times\Bbb Z^*_q$$

The above's based on the well known property of Euler's Totient Function

$$\phi(pq)=\phi(p)\phi(q)=(p-1)(q-1)$$

share|improve this answer
    
I know that both groups have the same order. However, since both p-1 and q-1 may not be prime, may your hint be incorrect? For example, if p=3, q=5, the group in the left has order 8, which is the order of $\mathbb{Z}_8$, different from $\mathbb{Z}_2$ X $\mathbb{Z}_4 $, –  NH.Jeong Oct 14 '13 at 15:44
    
The group $\;\Bbb Z_{15}^*\;$ is not cyclic...! –  DonAntonio Oct 14 '13 at 15:55
    
Oh I checked that they are isonorphic. You are right. Then is the answer of the second problem is the klein 4group because $\mathbb{Z}_{pq}^*$ is not cyclic? –  NH.Jeong Oct 14 '13 at 16:23
    
How can that be the Klein group if $\;p,q\;$ different odd primes?? –  DonAntonio Oct 14 '13 at 16:26
    
Um.. I applied the argument in my question, so I thought the order of $(\Bbb Z^*_p\times\Bbb Z^*_q)^2$ is the 1/4 times of that of $\Bbb Z^*_p\times\Bbb Z^*_q$ –  NH.Jeong Oct 14 '13 at 16:34

$\mathbb{Z}_{pq}^\ast = \mathbb{Z}_p^\ast \times \mathbb{Z}_q^\ast$, and $(\mathbb{Z}_{pq}^\ast)^2 = (\mathbb{Z}_p^\ast)^2 \times (\mathbb{Z}_q^\ast)^2$.

share|improve this answer
    
Hence is the answer is klein group? –  NH.Jeong Oct 16 '13 at 8:17
    
@JeongNam-ho That should be easy enough to check, no? –  you-sir-33433 Oct 16 '13 at 10:00
    
What are you saying..?T.T. So, Is it the klein group?? –  NH.Jeong Oct 16 '13 at 11:16
    
@JeongNam-ho Well, what is the answer for $n=15$? How would a confident mathematician ask this question? –  you-sir-33433 Oct 16 '13 at 11:27
    
I think, the answer is the klein group for the case n=15..... –  NH.Jeong Oct 16 '13 at 12:23

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.