Lesson 27 Solution

Exercise 1) Find 3 non-trivial subgroups of the group $\{0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11\}$ with addition modulo 12.

Solution: To be non-trivial, our subgroups cannot be the subset $\{0\}$ , 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) $\{0, 2, 4, 6, 8, 10\}$ ,

2) $\{0, 3, 6, 9\}$ ,

3) $\{0, 4, 8\}$

4) $\{0, 6\}$ .

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 $6$ is $6$ and the inverse of $9$ is $3$ . 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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On to Lesson 28

3 Responses to Lesson 27 Solution

yoel halb says:

February 22, 2015 at 2:06 pm

What about {0, 1, 11}?

- yoel halb says:
  
  February 22, 2015 at 3:41 pm
  
  Actually the same should hold true for any n1 and n2 such as n1+n2 = identity
  
  - TrueBeautyOfMath says:
    
    March 1, 2015 at 7:37 pm
    
    Unfortunately this isn’t quite the case. Recall that a subgroup must have all possible 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 🙂