# Lesson 27 Solution

**Exercise 1) ****Find 3 non-trivial subgroups of the group with addition modulo 12.**

Solution: To be non-trivial, our subgroups cannot be the subset , nor can they be the whole group. It turns out there are four non-trivial subgroups, so even though I’ve only asked for three (for some unknown reason), I’ll provide all four here. They are:

1) ,

2) ,

3)

4) .

All of these subsets have the three properties of being closed under addition, containing the identity, and containing all inverses. For example, in 2), the inverse of is and the inverse of is . I encourage the reader to check that all of the other subgroups satisfy all of these requirements, and then also to explore subgroups of other sets with modular arithmetic defined on them.

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What about {0, 1, 11}?

Actually the same should hold true for any n1 and n2 such as n1+n2 = identity

Unfortunately this isn’t quite the case. Recall that a subgroup must have

allpossible group multiplications in them. Thus, if we include 1 in the subgroup, then we must also include 1+1, which is 2. But then we also need to include 2+1, which is 3. We quickly see, then, that any subgroup that includes 1 must actually be the entire group! This is why we must be a bit more careful when defining subgroups 🙂