2012-03-13

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).

just another guy trying to make the diagrams commute.
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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.

• mlbaker says:

Thanks! That’s insightful.

4. nigelvr says:

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

• mlbaker says:

yeah, $\hookrightarrow$ is usually used for “monic”, $\twoheadrightarrow$ for “epi”