Is $(\mathbb{Z}/15\mathbb{Z})^\times$, the group of units of the ring $\mathbb{Z}/15\mathbb{Z}$, isomorphic to $(\mathbb{Z}/20\mathbb{Z})^\times$?

Question

Is $(\mathbb{Z}/15\mathbb{Z})^\times$, the group of units of the ring $\mathbb{Z}/15\mathbb{Z}$, isomorphic to $(\mathbb{Z}/20\mathbb{Z})^\times$?

Is $(\mathbb{Z}/5\mathbb{Z})^\times$ isomorphic to $(\mathbb{Z}/12\mathbb{Z})^\times$?

My attempt: I know that $(\mathbb{Z}/15\mathbb{Z})^\times$ and $(\mathbb{Z}/20\mathbb{Z})^\times$ both contain $8$ elements, so they are not immediately excluded. Is it true that $(\mathbb{Z}/15\mathbb{Z})^\times \cong (\mathbb{Z}/5\mathbb{Z})^\times \times (\mathbb{Z}/3\mathbb{Z})^\times$ and $(\mathbb{Z}/20\mathbb{Z})^\times \cong (\mathbb{Z}/5\mathbb{Z})^\times \times (\mathbb{Z}/4\mathbb{Z})^\times$? Can I use this to help answer the problem?

I believe $(\mathbb{Z}/5\mathbb{Z})^\times$ is not isomorphic to $(\mathbb{Z}/12\mathbb{Z})^\times$ because though they have the same number of elements, the latter has an element of order 11, while the former cannot possibly have an element of order 11.

Any help appreciated!

Answer 1

By the Chinese remainder theorem:

$ (\mathbb{Z}/15\mathbb{Z})^\times \cong (\mathbb{Z}/3\mathbb{Z})^\times \times (\mathbb{Z}/5\mathbb{Z})^\times \cong C_2 \times C_4 $

$ (\mathbb{Z}/20\mathbb{Z})^\times \cong (\mathbb{Z}/4\mathbb{Z})^\times \times (\mathbb{Z}/5\mathbb{Z})^\times \cong C_2 \times C_4 $

$C_n$ is the cyclic group of order $n$.

Answer 2

Generally if $A, B$ are rings then $(A\times B)^{\times}$ is isomorphic to $A^{\times}\times B^{\times}$. That's quite simple to prove. You just have to realize that $(a,b)$ is invertible if and only if both $a,b$ are invertible in their respective rings.

Now it is well known that if $n=p_1^{a_1}\cdots p_n^{a_n}$ for distinct primes $p_i$ then

$$\mathbb{Z}_{n}\simeq \mathbb{Z}_{p_1^{a_1}}\times\cdots\times\mathbb{Z}_{p_n^{a_n}}$$

as rings. I'm using simplified notation $\mathbb{Z}_n=\mathbb{Z}/n\mathbb{Z}$. So in order to find the group of units for a cyclic ring all you have to do is to consider prime powers. Now describing it generally can be tricky. Here's the result:

$$(\mathbb{Z}_2)^{\times}\simeq 0$$ $$(\mathbb{Z}_4)^{\times}\simeq \mathbb{Z}_2$$ $$(\mathbb{Z}_{2^k})^{\times}\simeq\mathbb{Z}_2\times\mathbb{Z}_{2^{k-2}}\mbox{ for }k\geq 3$$ $$(\mathbb{Z}_{p^k})^{\times}\simeq\mathbb{Z}_{p^k - p^{k-1}}\mbox{ for prime }p\neq 2$$

I'm not going to give you full proof of that. Cause I don't know how to prove it myself and the only resource I found it in was some old text book. I guess you have to believe me. :D

Combining all those pieces of information should give you precise answer to your question and all other possible in this context.