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.

For two distinct primes $q$ and $p$, is there non-trivial homomorphism from $\mathbb{Z}_q^*$ to $\mathbb{Z}_p^*$? Here, $\mathbb{Z}_q^*$ and $\mathbb{Z}_p^*$ mean the multiplication groups with respect to modulus q and p, respectively, while "non-trivial" means we do not want the trivial homomorphism that maps all elements to 1.

share|improve this question

1 Answer 1

If $\;p,q\;$ are odd primes then their multiplicative groups are cyclic of even order and thus both groups have a cyclic subgroup of order two...

For example

$$\phi: \Bbb Z_5^*=\langle 2_5\rangle\to\Bbb Z_{11}^*=\langle 2_{11}\rangle\;,\;\;\phi(2_5):=2_{11}^5\implies$$

$$\phi(2_5^2=4)=2_{11}^{10}=1\;,\;\;etc.$$

The symbol $\,2_p\,$ means the element two modulo $\,p\,$ (in both cases above it is a generator of the mutiplicative group. This is not always the case...)

share|improve this answer
    
Well! Thanks! Furthermore, what if I want a homomorphism between large subgroups of $\mathbb{Z}_q^*$ and $\mathbb{Z}_p^*$? The subgroup of order two is too small to my on-going project. Thank you again! –  Pigmann Jun 17 '13 at 10:50
2  
Well, then you have to check particular cases. For example, if both primes are $\,1\pmod 4\,$ then there is a subgroup in each of order $\,4\,$ ...Even if you have more or less large primes, say $\,p=4001\;,\;\;q=5501\;$ , the corresponding multip. groups' orders are $\,4000=2^5\cdot 5^3\;,\;5500=2^2\cdot 5^3\cdot 11\,$ , you can find isomorphic subgroups of rather limited orders ($\,2^2\;,\;\;5^3\;$ , say) –  DonAntonio Jun 17 '13 at 11:02
    
Great! Thanks a lot! –  Pigmann Jun 17 '13 at 11:08

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.