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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...

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    $\begingroup$ What is $U(33)$? $\endgroup$ – José Carlos Santos Feb 10 '18 at 22:11
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    $\begingroup$ $U(33)$ is $(\mathbb{Z}/33\mathbb{Z})^{\times}$? $\endgroup$ – Delong Feb 10 '18 at 22:14
  • $\begingroup$ 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}. $\endgroup$ – JSAlg Feb 10 '18 at 22:35
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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.

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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) .

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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 $U(33)\cong H\times K\cong \mathbb Z_2\times\mathbb Z_2\times\mathbb Z_5\cong\mathbb Z_2\times\mathbb Z_{10}$.

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  • $\begingroup$ 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 $2$, $U(33)=Z_2\times Z_{10}$. $\endgroup$ – jgon Jun 13 at 23:42

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