Dinner and math.

EDIT: Here’s an interesting sidenote. If you try to generalize your approach to PMATH 346 A5P4 (e.g., try to count how many distinct isomorphism classes of finite abelian groups of order N there are — for arbitrary N) then you will find it is related to the number of integer partitions of the exponents of the primes appearing in the prime factorization of N.

Some random problems I thought up:

What is the automorphism group of a dihedral group? (So meta…)

Is there a group G such that the sequence defined by A_1 = \mathrm{Aut}(G), and A_n = \mathrm{Aut}(A_{n-1}) for all n \geq 2 never becomes constant?

Does there exist an infinite group G, a non-trivial inner automorphism \phi : G \to G (note immediately that this means G must be non-abelian if it exists), and proper infinite subgroups A and B so that \phi(A) \subseteq B but \phi(A) \neq B?

Are there two non-isomorphic groups G and H with injective homomorphisms G \hookrightarrow H and H \hookrightarrow G?

(Feel free to submit responses).


About mlbaker

just another guy trying to make the diagrams commute.
Aside | This entry was posted in articles. Bookmark the permalink.

7 Responses to 2012-03-13

  1. fomalhauty says:

    campus bubble! 😛

    • mlbaker says:

      I was pretty cynical of its ability to compare to Sweet Dreams… but I think I’m pretty satisfied after trying it (so begins the addiction) XD

  2. crobert says:

    Re: 3rd question. Take three nonabelian groups L, K, M and look at their direct product L \times K \times M. pick some a \in L (not equal to 1) and set \phi to be conjugation by (a,1,1). set A = L \times 1 \times 1, B = L \times K \times1. \phi(A) = A which is a proper subset of B.

  3. crobert says:

    Re: 4th question (this time for real)

    joint soln with chenglong

    let A = \mathbb{Q} \times \mathbb{Q} \times \ldots so countably many copies of \mathbb{Q}, and let B = \mathbb{Z} \times \mathbb{Q} \times \mathbb{Q} \times \ldots so countably many copies of \mathbb{Q} with a \mathbb{Z} stuck to the front. Clearly B embeds in A, but A also embeds in B by shifting everything over to the right. now we claim that there is no isomorphism from B to A. indeed, if there was such a map we just look at what it does to (1,0,0,\ldots), say it gets mapped to (a_0,a_1,a_2,\ldots). then the map is onto so there is something that gets mapped to (a_0/2, a_1/2, a_2/2, \ldots) which can’t come from \mathbb{Z} \times \{0 \} \times \{ 0 \} \times \ldots because there’s no 1/2 in \mathbb{Z}. but now if a maps to this thing then 2a maps to the same thing as (1,0,\ldots) which violates injectivity.

  4. nigelvr says:

    what does the arrow with the curved left end mean? Does it just mean the homomorphism is injective?

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