# Proving U(33) is isomorphic to $Z_{10}\oplus Z_2$?

I am trying to prove that the group $U(33)$ is isomorphic to $Z_{10}\oplus Z_2$, however I am struggling to find an equation $\phi:U(33)\rightarrow Z_{10}\oplus Z_2$ such that $\phi$ is a homomorphism and a bijection.

I know that both groups are abelian, and both have a group order of 20, but I don't know how exactly to come up with some function between the 2 that is an isomorphism. Where do I start? I noticed the cycles of many of the elements of $U(33)$ are of order 10, so I am wondering if I can use that in the equation.

Also, since $\phi$ has to be injective, I know $\phi(1) = (0,0)$. Not sure how to map everything else...

• What is $U(33)$? – José Carlos Santos Feb 10 '18 at 22:11
• $U(33)$ is $(\mathbb{Z}/33\mathbb{Z})^{\times}$? – Delong Feb 10 '18 at 22:14
• U(33) is the group of integers (in mod 33), who is relatively prime to 33. So {1, 2, 4, 5, 7, 8, 10, 13, 14, 16, 17, 19, 20, 23, 25, 26, 28, 29, 31, 32}. – JSAlg Feb 10 '18 at 22:35

Directly

$$\Bbb Z_{33}\cong\Bbb Z_{11}\oplus\Bbb Z_3\implies U\left(\Bbb Z_{33}\right)\cong U\left(\Bbb Z_{11}\right)\oplus U\left(\Bbb Z_3\right)=\Bbb Z_{10}\oplus\Bbb Z_2$$

Check, first of all, the first implication. That's all you need.

The homomorphism $$\phi:\Bbb Z\to\Bbb Z_{11}×Z_3$$ determined by $$\phi(1)=(1,1)$$ induces an isomorphism. It is clear that the kernel is $$33\Bbb Z$$.

Or, using Bezout's theorem, we can write $$11\cdot 5-3\cdot 18=1$$. Then $$\psi:\Bbb Z_{11}×\Bbb Z_3\to\Bbb Z_{33}$$ given by $$\psi((a,b))=55a-54b$$ is an isomorphism. For the kernel is $$\{(0,0)\}$$.

Now since $$\Bbb Z_{33}\cong\Bbb Z_{11}×\Bbb Z_3$$, we have $$(\Bbb Z_{33})^×\cong (\Bbb Z_{11}×\Bbb Z_3)^×$$.

Now for a little category theory. There is a natural functor $$U$$ from the category of rings ($$\bf{Ring}$$) to the category of groups ($$\bf{Grp}$$), that sends a ring $$R$$ to its group of units $$U(R)$$. This functor respects direct products, because it is a right adjoint functor (the left adjoint being the group ring construction) .

It seems that $H=\{1, 10, -1=32, -10=23\}$ is a subgroup of $U(33)$ isomorphic to $\mathbb Z_2\times\mathbb Z_2$. On the other hand, you have the subgroup $K=\langle 2\rangle=\{1,2,4,8,16\}$ of order $5$, which is isomorphic to $\mathbb Z_5$. Because they intersect only trivially, and $|H||K|=20=|U(33)|$ then $U(33)\cong H\times K\cong \mathbb Z_2\times\mathbb Z_2\times\mathbb Z_5\cong\mathbb Z_2\times\mathbb Z_{10}$.

• Doesn't $2$ have order $10$, since $2^5=-1$? Also then they don't intersect trivially. However, I think you can say $2$ has order $10$ and its cyclic subgroup intersects that generated by $10$ trivially. Since $10$ has order $2$, $U(33)=Z_2\times Z_{10}$. – jgon Jun 13 at 23:42