Are the groups $(\mathbb{Z}/n\mathbb{Z})^{\times}$ and $(\mathbb{Z}/m\mathbb{Z})^{\times}$ isomorphic if they have the same order? Let $(\mathbb{Z}/n\mathbb{Z})^{\times}$ the set of units of $\mathbb{Z}/n\mathbb{Z}$ (i.e the elements with multiplicative inverse). And consider the groups:
$$(\mathbb{Z}/17\mathbb{Z})^{\times}, (\mathbb{Z}/32\mathbb{Z})^{\times}, (\mathbb{Z}/34\mathbb{Z})^{\times}, (\mathbb{Z}/40\mathbb{Z})^{\times}, (\mathbb{Z}/48 \mathbb{Z})^{\times}$$
Show that at least 2 are isomorphic. (Hint: Verify that they have the same order. It's not necessary to determine which two groups are isomorphic).
So, following the hint, I calculated the order of the groups and found that
$\vert (\mathbb{Z}/17\mathbb{Z})^{\times} \vert= 17, \vert (\mathbb{Z}/32\mathbb{Z})^{\times} \vert= 16, \vert (\mathbb{Z}/34\mathbb{Z})^{\times} \vert= 16, \vert (\mathbb{Z}/40\mathbb{Z})^{\times} \vert= 15, \vert (\mathbb{Z}/48\mathbb{Z})^{\times} \vert= 8.$
Therefore, the candidates are $(\mathbb{Z}/32\mathbb{Z})^{\times}$ and $(\mathbb{Z}/34\mathbb{Z})^{\times}$. However when looking at the elements, both of the groups are formed by the equivalence classes of odd numbers, but $(\mathbb{Z}/32\mathbb{Z})^{\times}$ includes the class $\overline{17}$, while $(\mathbb{Z}/34\mathbb{Z})^{\times}$ doesn't have $\overline{17}$ but instead $\overline{33}$. Moreover, I know that the common equivalence classes won't have the same order since for the first, the classes are $\text{mod}(32)$ and for the second, they are $\text{mod}(34)$. So I discarded the idea of setting a map from each class in $(\mathbb{Z}/32\mathbb{Z})^{\times}$ to the same one in $(\mathbb{Z}/34\mathbb{Z})^{\times}$, $\overline{17} \mapsto \overline{33}$ and showing it is a homomorphism.
So, my question is, from the hint it kinda looks like it is enought to show they have the same order to conclude they are isomorphic. But I don't see why and even more, I wouldn't know how to show that they are isomorphic without narrowing down the options to the two candidates.
 A: The given groups are $ U_{17}, U_{32}, U_{34}, U_{40} \, and \, U_{48} $ which are respectively isomorphic to
$\mathbb Z_{16}, \mathbb Z_ 2 \times \mathbb Z_8 , U_2 \times U_{17}, U_8\times U_5 \, and \, U_{16} \times U_3$. These are respectively isomorphic to $\mathbb Z_{16}, \mathbb Z_2 \times \mathbb Z_8, \mathbb Z_{16}, \mathbb Z_2 \times \mathbb Z_2 \times \mathbb Z_4, \, and \, \mathbb Z_2 \times \mathbb Z_4\times \mathbb Z_2 $. Hence the first and the third groups are isomorphic and also the fourth and fifth groups are also isomorphic. I have used the isomorphism theorems pertaining to the multiplicative groups "$ U_{pq} \, is \, isomorphic \, to \, U_p \times U_q $" when p and q are coprime and also results relating to converting multiplicative groups of orders with even power prime and odd power primes into product of cyclic groups.
A: You can see fairly directly that at least $(\mathbb{Z}/17\mathbb{Z})^{\times}$ and $(\mathbb{Z}/34\mathbb{Z})^{\times}$ are isomorphic.
Indeed, $(\mathbb{Z}/34\mathbb{Z})^{\times}$ are the set of integers
$\{k \in \{1,2, \ldots, 33\}$; $k$ odd;$ (k,17)=1\}$. So let $\phi$: $(\mathbb{Z}/34\mathbb{Z})^{\times} \mapsto (\mathbb{Z}/17\mathbb{Z})^{\times}$; be the following mapping:
$$\phi(k) = k \mod 17$$
Then $\phi$ is indeed an isomorphism. You can also check the following: $\phi^{-1}$: $(\mathbb{Z}/17\mathbb{Z})^{\times} \mapsto (\mathbb{Z}/34\mathbb{Z})^{\times}$; is as follows: $\phi^{-1}(\ell)$
$=$ $\ell +17$ iff $\ell \in \{2,4,6,8, \ldots, 16\}$; $\phi^{-1}(\ell) = \ell$ iff $\ell \in \{1,3, \ldots, 15\}$.
In fact this trick can be extended to show
$(\mathbb{Z}/M\mathbb{Z})^{\times}$ is isomorphic to $(\mathbb{Z}/2M\mathbb{Z})^{\times}$ for any odd $M \ge 3$.
However, already noted in the comments I believe, to answer the title question of this thread,$(\mathbb{Z}/17\mathbb{Z})^{\times}$ has 8 squares whereas $(\mathbb{Z}/32\mathbb{Z})^{\times}$ has only at most 4, namely $k \equiv_8 1$, so even though they have the same order, they are not isomorphic.
