# Is it true that if all elements of a group (multiplication modulo $n$) are composed with any one element, then they become another group?

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?

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 $$1\in A$$.