What's a multiplicative group?
By Sophia Bowman
What's the multiplicative group of integers modulo $n$? I checked out a couple of websites, including Wikipedia, and all I understood was they are the modulo $n$ integers coprime to $n$.
Is that true? Is it simply the set of modulo $n$ integers comprime to $n$? Does this set have any unique qualities?
$\endgroup$ 12 Answers
$\begingroup$By definition a group is a set, together with a binary operation $*$ which is associative, has an identity element, and for which every element has an inverse. For example, the integers with the operation $+$ form a group. The identity is 0, the inverse is the opposite. However, the integers with the operation of multiplication don't form a group (what is the identity element? OK, fine, it's 1, but now what is the inverse of 2?). The rational numbers still don't form a group under multiplication, but the rational numbers except for 0 do (the identity is 1, the inverse is the reciprocal).
Back to your question - the underlying set of the multiplicative group is the set of integers coprime to $n$, but the operation is multiplication modulo $n$. It is worth thinking about why this is actually a group operation - there's clearly an identity (1), but why are there inverses?
EDIT: I assume you are looking to understand the answers to your questionExplanation of a mathematical phenomenon?more thoroughly. Even though the question can be answered without group theory, it's a good motivation to learn group theory since there are many similar questions for which elementary techniques aren't available.
It indeed relates to the multiplicative group of units mod $n$ (in the case that $n$ is prime). The relevant properties of this group for your question is that it has $p - 1$ elements in it, and that the sets you consider in the other question are cosets of a particular subgroup. You will need Lagrange's theorem to get the answer to your earlier question - actually, the proof of Lagrange's theorem is the same as the "elementary" proof of your earlier question.
$\endgroup$ 1 $\begingroup$I hope you know ($\mathbb{Z_n},+)$, now take the elements which are less than equal to n and prime to n and give a multiplication that set like [a][b]=[ab] for all [a],[b] in $\mathbb{Z_n}$ check that this is a binary operation on $\mathbb{Z_n}$ and associative. Now [1] is in $\mathbb{Z_n}$ and for all [a] in $\mathbb{Z_n}$ [a][1]=[1]=[1][a] so identity is there, and let [a] is in $\mathbb{Z_n}$ and gcd(a,n)=1 then find that there exist b such that [a][b]=[b][a]=1 (apply some number theory elementary), let $U_n$ be the set of all elements [a] in $\mathbb{Z_n}$ such that $$U_n=\{[a]\in \mathbb{Z_n}\setminus \{0\}:(a,n)=1\}$$ which is multiplicative modulo n, group, for n=8 $U_8=\{[1][3][5][7]\}$
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