# Is there non-trivial homomorphism from $\mathbb{Z}_q^*$ to $\mathbb{Z}_p^*$?

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.

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## 1 Answer

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...)

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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
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