Considering the binary structure <{4, 8, 12, 16}, mult. mod 20>, I have to prove that it is a group. I have figured out that the identity is 16 and found that the inverses exist, but I'm not sure how to prove associativity and "closed-ness" efficiently. The only idea I have for associativity is to just brute-force it and show every multiplication, but I don't feel that is a good way to learn. So I guess my questions are:

1) Are all sets associative under multiplication modulo n?

2) Are all sets closed under multiplication modulo n?

I feel the answers are yes to both since mult. modulo n means all elements are "contained" within {0, ... , n-1}, but I don't know how to prove (or disprove) it. Any help would be great. Thanks!

  • $\begingroup$ I assume instead of 1, ...,n-1 you meant 0,...,n-1. $\endgroup$
    – quid
    Sep 25, 2016 at 13:10
  • $\begingroup$ Oops, you're right, edited it to fix it. $\endgroup$
    – Max
    Sep 25, 2016 at 13:15

2 Answers 2


Re 1: Yes. But it is not the set that is associative, but multiplication of elements from the set, or multiplication over the set if you prefer. You thus do not need to prove it for this case specifically. Just recall that multiplication $\mod n$ for any $n$ is associative and this property is thus inherited when it is restricted to your subset.

Re 2: No. Consider the set $\{2, 16\}$ modulo $20$. You do not have that $2 \times 16$ is in the set as it is $12$ modulo $20$ and $12$ is not in your set. I think there brute force is the intended method. Just check all products. (Do not forget those of an element with itself.) It is true that the product is still some class modulo $20$ but this is not what it means that a set is closed under the operation. The result most be an element from the set in which your started.


For associativity, note that multiplication modulo any number is always associative.

For closed-ness, note that the product of any two numbers in $\{4,8,12,16\}$ will remain a multiple $4$ and cannot become a multiple of $5$.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .