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.

Back to the lesson

On to Lesson 28

Advertisements

3 Responses to Lesson 27 Solution

  1. yoel halb says:

    What about {0, 1, 11}?

    • yoel halb says:

      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 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 🙂

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s