I read the following question in Gallian's Contemporary Abstract Algebra:

Show that the set $A = \{5, 15, 25, 35\}$ is a group under multiplication modulo 40. What is the identity element of this group? Can you see any relationship between this group and $B = U(8)$? ($U(8)$ is the set of all positive less than 8, and coprime to it; group operation is multiplication modulo 8)

The first part of the question was easy. Now, for the second part of the question, I noticed that all the elements of the group A are just the elements of $B=\{1,3,5,7\}$ multiplication modulus 8, multiplied with five, with the modulus value equal to $40=8\times5$ too. I tried this for another group ($U(12)$), and it worked then too.

I now generalized it as follows:

Given a group $A=U(n)=\{1, a_1, a_2, ..., a_p\}$ for $n>1$, the set $B=\{a_i, a_i\times a_1, a_i\times a_2, ..., a_i\times a_p\}$ is also a group, multiplication modulo $n\times a_i$, for any integer $i\in [1,p]$.

I tried to prove it by showing that there exists, in the second set: an identity , an inverse for every element, and that the operation is associative.

For example, assume in $A$ the identity element was $a_m$, then $a_r\times a_m\equiv a_r \text{ (mod $n$)}$ (for any integer $r$ in $[1, n]$), or $a_r\times a_m=a_r + nk$. Now, for the second set, by observation I assume that identity element would be $a_i\times a_m$. Then, $(a_i\times a_r)\times (a_i\times a_m)=(a_i\times a_r) + (n\times a_i)\times k'$ should be true if it is a group. This gives us $a_i\times (a_r\times a_m)=a_r + nk'$. Putting $a_r\times a_m=a_r + nk$, we get $a_i\times (a_r+nk)\overset{??}{=}a_r + nk'$, which is a statement I am not sure if it's correct or not.

Thus, I am stuck here, and unable to prove my generalization. I have thought of many sample cases for different values of $n$, and they all seem to be working. So, I am pretty sure my generalization should be correct. What steps am I missing here? Or is my generalization false? If so, what's the counterexample?


1 Answer 1


Let $m$ be coprime to $n$. Then $A:=m U(n)$ is a group under multiplication modulo $mn$, and $A\approx U(n)$.

Indeed, $A$ is closed under multiplication (if $m\mid x$ and $m\mid y$, then $m\mid xy$ and $m\mid xy\bmod{mn}$). Consider he map $\phi\colon A\to U(n)$, $x\mapsto x\bmod n$ (note that this can be defined because if $x=my\in m U(n)$, then $\gcd(x,n)=\gcd(m,n)\gcd(y,n)=1$).

  • Then clearly $\phi(xy)=\phi(x)\phi(y)$.

  • Also, $\phi$ is injective (because $\phi(x)=\phi(y)$ implies $x\equiv y\pmod n$ and together with $x\equiv y\pmod m$ we get $x\equiv y\pmod {nm}$)

  • As $A$ and $U(n)$ are finite sets of the same cardinality, this implies that $\phi$ is a bijection

We conclude that $\phi$ is a magma isomorphism between $A$ and $U(n)$, and as $U(n)$ is a group, this meams that the magma $A$ is in fact a group and $\phi$ is a group isomorphism. $\square$

It may be worth noting that the neutral element of $A$ is not $m\cdot 1$, but rather the solution to $x\equiv 1\pmod n$, $x\equiv 0\pmod m$ that exists by virtue o fthe Chinese Remainder Theorem. (In your example, $25$)

The condition that $m$ and $n$ are coprime is necessary: If $d=\gcd(m,n)$, then the product of any two elements of $A$ is a multiple of $d^2$ (even after reduction modulo $mn=d^2\cdot \frac md\frac nd$), hence can never be $1\in A$.


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