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Suppose $U(n)$ is the multiplicative group of units of $\mathbb Z_n$. We know that $U(m)$ and $U(n)$ may be isomorphic even if $m\neq n$, for example $U(8)\cong U(12)$. So,

I am looking for a necessary and sufficient condition on $m,n\in \mathbb N$ for $U(m)\cong U(n)$.

Can someone provide me with such a condition? I am new in group theory and so I cannot think of a suitable condition. Can someone please help?

Nothing special
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Kishalay Sarkar
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3 Answers3

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It's necessary that $\varphi(m)=\varphi(n)$. But it is not sufficient. For instance, $\varphi(21)=\varphi(13)$. But $U(21)\not\cong U(13)$. The latter is cyclic, the former is not.

At the moment, I'm not sure about a sufficient condition.

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As the other answer suggested, it is necessary that $\varphi(m)=\varphi(n)$ for $U(m)$ and $U(n)$ to be isomorphic, however, it is not sufficient.

I am making an attempt at finding a sufficient condition.

Theorem: $U(m_1 m_2)\cong U(m_1)\times U(m_2)\tag*{}$ if and only if $\text{gcd}(m_1,m_2)=1$.

Using induction, you may as well extend this theorem.

Theorem 1: $U(m_1m_2\cdots m_k)\cong U(m_1)\times U(m_2)\times \ldots \times U(m_k)\tag*{}$ if and only if $\text{gcd}(m_i, m_j)=1$ whenever $i\neq j$.

Theorem 2: $U(n)$ is cyclic i.e., $U(n)\cong C_{\varphi(n)}\tag*{}$ if and only if $n$ is of the form $1$, $2$, $4$, $p^k$ or $2p^k$ where $p$ is an odd prime and $k\in \mathbb Z^+$.

Theorem 3: $U(2^n)\cong C_{2^1}\times C_{2^{n-2}}\tag*{}$ for all integers $n\geq 2$.

With this in mind, if you are given two numbers $m$ and $n$, in most of the cases, you can factorise $U(m)$ and $U(n)$ as direct product of cyclic groups using these theorems.

Example 1: $\varphi(8)=\varphi(10)=\varphi(12)=4$

$8=2^3$ (use theorem 3), $10=2\times 5$ (use theorem 2) and $12=2^2\times 3$ (use theorem 1 then 3 and 2).

$U(8)\cong C_2\times C_2\tag*{}$

$U(10)\cong C_4\tag*{}$

$U(12)\cong U(4)\times U(3) \cong C_2\times C_2\tag*{}$

Hence, $U(8)\cong U(12)\not \cong U(10)$.

Example 2: $\varphi(41)=\varphi(100)=40$

$U(41)\cong C_{40}\tag*{}$

$U(100)\cong U(4)\times U(5^2) \cong C_2\times C_{20}\tag*{}$

Recall the Chinese remainder theorem, $\mathbb Z_m\times \mathbb Z_n\cong \mathbb Z_{mn}$ if an only if $\text{gcd}(m,n)=1$.

Hence, $U(41)\not \cong U(100)$.

Example 3: $\varphi(44)=\varphi(33)=20$

$U(44)\cong U(2^2)\times U(11)\cong C_{2}\times C_{10}\tag*{}$

$U(33)\cong U(3)\times U(11) \cong C_2\times C_{10}\tag*{}$

Hence, $U(44) \cong U(33)$.

Example 4: $\varphi(64)=\varphi(68)=32$

$U(64)\cong C_{2}\times C_{2^4}\tag*{}$

$U(68)\cong U(4)\times U(17) \cong C_2\times C_{16}\tag*{}$

Hence, $U(64) \cong U(68)$.

Example 5: $\varphi(70)=\varphi(45)=24$

$U(70)\cong U(2)\times U(5)\times U(7)\cong C_{1}\times C_{4}\times C_{6}\tag*{}$

$U(45)\cong U(3^2)\times U(5) \cong C_{6}\times C_{4}\tag*{}$

Hence, $U(70) \cong U(45)$.

Example 6: $\varphi(69)=\varphi(92)=44$

$U(69)\cong U(3)\times U(23) \cong C_{2}\times C_{22}\tag*{}$

$U(92)\cong U(4)\times U(23) \cong C_{2}\times C_{22}\tag*{}$

Hence, $U(69) \cong U(92)$.

I will update this answer with more rigorous step by step procedure, if possible.

Nothing special
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  • To avoid confusion with the ring $\mathbb Z_n$, I suggest using $C_n$ for the cyclic group of order $n$. – lhf Oct 30 '23 at 10:46
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    @lhf Ok, I will make the changes... Most study materials and textbooks use $\mathbb Z_n$ to refer to the cyclic group of order $n$ where I live. – Nothing special Oct 30 '23 at 12:55
  • Thanks. It’s common to conflate an abstract group with its concrete realizations. – lhf Oct 30 '23 at 13:56
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    I doubt there is any simple necessary and sufficient condition. (We don't want the useless answer "they are isomorphic if and only if they are isomorphic".) When $p$ is an odd prime dividing $m$, $U(m)$ has a direct product decomposition involving a cyclic component of order $p-1$, which in a direct product decomposition breaks up more (unless $p$ is $3$ or another Fermat prime) and this seems hard to control. Consider $\varphi(43) = \varphi(49) = 42$: $U(43) \cong C_{42} \cong C_2 \times C_3 \times C_7$ and $U(49) \cong C_6 \times C_7 \cong C_2 \times C_3 \times C_7$. – KCd Oct 30 '23 at 14:01
  • @KCd I agree, to say that they are isomorphic iff they decompose the same way is to say that they are isomorphic iff they are isomorphic... That's why I didn't say this verbatim :/ I just intended to add these theorems which other answers haven't... I would see that $43$ is prime or that $49$ is the square of an odd prime and conclude that both $U(43)$ and $U(49)$ are cyclic (from the theorem 2)... – Nothing special Oct 30 '23 at 20:56
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Firstly notice that the groups $U_n$ and $U_m$ are finite. So to be isomorphic their order have to be equal. Moreover isomorphism maps generator to generator. So from this you can guess that $\phi{(n)}=\phi{(m)}$. I hope you got the idea.

Edit : In that case you can prove $f: U_n \to U_m$ defined by $f(a)=a'$ is an isomorphism where $a$ and $a'$ are the generators of the cyclic groups $U_n$ and $U_m$ respectively. In this case the above condition is necessary and sufficient.

Itachi
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