# Abstract Nonsense

## R and C are Isomorphic as Groups

Point of Post: In this post we prove that $\mathbb{R}$ and $\mathbb{C}$ are isomorphic as abelian groups, as well as discussing a slightly more general theorem concerning when infinite abelian groups are isomorphic to finite powers of themselves.

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Motivation

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I rarely post about random theorems/facts/ideas that come up in my daily mathematical life. I think this type of post is actually very helpful since, from a blog-reader point of view, these are often the most entertaining posts. To this end I’d like to discuss a problem that came up in discussion with a friend today. The problem actually came up when we were discussing a different, still interesting (I hope to discuss it on here), question altogether. In particular, say a category $\mathcal{C}$ has the Schroeder-Bernstein property if whenever $x$ and $y$ are objects of $\mathcal{C}$ such that there exists monomorphisms $x\overset{\displaystyle \longleftarrow}{\longrightarrow}y$ then $x\cong y$. The classic Schroeder-Bernstein theorem then reduces to the statement that $\mathbf{Set}$ enjoys the Schroeder-Bernstein property. A natural question then comes as to which of the common categories have the Schroeder-Bernstein property. A quick check shows that $\mathbf{FinGrp}$, $\mathbf{FinRing}$, and $\mathbf{FinVect}_k$ all do, where the first two categories are finite groups and finite rings respectively. Ok, so ‘finite’ things tend to have the Schroeder-Bernstein property, so what are some examples of categories that don’t? I was aware of the fact that $\mathbf{Ab}$ did not have the Schroeder-Bernstein property but that the category of finitely generated abelian groups did.  So, me and my friend set out to find the examples in $\mathbf{Ab}$ which would violate the Schroeder-Bernstein property. So, being the good little math students that we are, we jumped right at the first non-finitely generated abelian group that came to mind, $\mathbb{R}$. Then, looking for another non-finitely generated abelian group (besides $\mathbb{R}$) to throw $\mathbb{R}$ against we immediately jumped at $\mathbb{C}$. There is an obvious embedding $\mathbb{R}\hookrightarrow\mathbb{C}$, and we started to wonder about embeddings the other way, when the obvious thing hit us. This is all moot if $\mathbb{R}\cong\mathbb{C}$ as abelian groups? Of course, such an ‘obvious’ question should have an obvious answer, right? Well, embarrassing as it is, neither of us knew the answer off hand. It took me the rest of walk to the car, and half the ride home to figure out the answer–and a simple answer it is–yes, they are isomorphic. The proof is actually a one-liner. This is just a reminder to always ask the ‘obvious’ questions, because sometimes we forget to.

January 25, 2012

## Homomorphisms Between Finitely Generated Abelian Groups (Pt. I)

Point of Post: In this post we derive the necessary information to describe $\text{Hom}_\mathbb{Z}(A,B)$ for any finitely generated abelian groups $A,B$ according to their cyclic group decomposition.

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Motivation

In this post, using our ability to split Hom across finite products we shall show how to effectively calculate the Hom group between any two finitely generated abelian groups. The only caveat to this is that one must first decompose the finitely generated abelian groups into their cyclic decomposition as is guaranteed by the structure theorem. Probably the most useful aspect of this post is that it will, if you keep your bookkeeping orderly, the number of homomorphisms between two finite abelian groups (presented as a product of cyclic groups) and the generators that will enable one to, in theory, produce every single homomorphism.

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November 14, 2011

## Unit Group of a Finite Field is Cyclic

Point of Post: In this post we prove the basic fact that if $k$ is a finite field then $U(k)$ is cyclic.

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Motivation

A very useful fact in both field and group theory is that if $k$ is a finite field then the group of units $U(k)$ is cyclic of order $|k|-1$. There are several ways one may go about doing this. In particular, there are two striking group theoretic facts about $U(k)$ which one hope characterize it. Namely, from basic field theory we know for any polynomial $p(x)\in k[x]$ one has that the number of roots of $p(x)$ in $k$ is bounded above by $\deg(p(x))$. In particular, using only the operation of the group we have that $x^n=1$ has at most solutions in $k$. The other interesting fact, which is easily implied by the bounded amount of roots $x^n=1$, is that $U(k)$ has at most one cyclic subgroup of a given order (this is clearly implied by the previous property since if $|g|=n$ and $C_n$ is the cyclic subgroup generated by $g$ then by Lagrange’s Theorem $x^n=x^{|C_n|}=1$ for each $x\in C_n$ and so, in particular $C_n$ constitutes $n$ solutions of $x^n=1$ and so clearly there can’t be another distinct cyclic subgroup of order $n$). So, we prove that both of these properties characterize cyclic groups and so, in particular, $U(k)$ is cyclic.

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September 21, 2011

## Sylow’s Theorems Revisited

Point of Post: In this post we give a more refined proof of Sylow’s Theorems.

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Motivation

On this blog I have given a proof of Sylow’s theorems and an alternate proof of Sylow’s first theorem. A year or so later, with more time, and more finesse I’d like to give the simplest, most coherent proof of Sylow’s theorems.

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September 20, 2011

## A Clever Proof of a Common Fact

Point of Post: In this post we give a new proof that if $G$ is a finite group $H$ is a subgroup $G$ whose index is the smallest prime dividing $|G|$ then that subgroup is normal.

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Motivation

It is a commonly used theorem in finite group theory that if $G$ is a finite group and $H\leqslant G$ such that $\left[G:H\right]=p$ is the smallest prime dividing $|G|$ then $H\unlhd G$. We have already seen a proof of this fact by considering the homomorphism $G\to S_p$ which is the induced map from $G$ acting on $G/H$ by left multiplication, and proving that $\ker(G\to S_p)=H$. We now give an even shorter (and the just mentioned proof is already short) proof of this fact using double cosets.

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September 16, 2011

## Actions by p-Groups (Pt. II)

Point of Post: This is a continuation of this post.

September 15, 2011

## Actions by p-Groups (Pt. I)

Point of Post: In this post we discuss the theory of $p$-groups acting on sets, and some of its ramifications.

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Motivation

I have previously discussed group actions, but being a rush to discuss group theory I skirted over some of the beautiful theory. So, I’d like to take some time to discuss one of the prettier and more powerful branches of the theory, namely when we restrict our attention to group actions by $p$-groups. Not only will we be able to say some prove some fairly substantive  theorems about $p$-group actions explicitly, but will be able to prove some very neat things in more general group theory and in number theory. The interesting fact about the theory we will discuss is that at the root of everything is a ‘fundamental theorem’ whose presence (being the theorem in the case of a particular group action) is the proof that every $p$-group has a non-trivial center.  Namely, we were able to conclude that since the cardinality of any conjugacy class must divide the order of the group, that they must be divisible by $p$. From this and the fact that the sum of the cardinalities of all the distinct conjugacy classes must sum to the order of the group (which is divisible by $p$) that the sum of all the one point conjugacy classes must have cardinality divisible by $p$. Well, the generalization of this idea (which, as I’m sure is pretty clear, can be restated for an arbitrary action with conjugacy class replaced by orbit) will be the main tool I mentioned from which all our other theorems are (not always straight-forward) consequences.

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September 15, 2011

## A Classification of Integers n for Which the Only Groups of Order n are Cyclic (Pt. II)

Point of Post: This is a continuation of this post.

September 13, 2011

## A Classification of Integers n for Which the Only Groups of Order n are Cyclic (Pt. I)

Point of Post: In this post we conglomerate and extend a few exercises in Dummit and Foote’s Abstract Algebra which will prove that the only positive integers $n$ for which the only group (up to isomorphism) of order $n$  is $\mathbb{Z}_n$ are integers of the form $n=p_1\cdots p_m$ are distinct primes with $p_i\not\equiv 1\text{ mod }p_j$ for any $i,j\in[m]$.

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Motivation

This post will complete several lemmas/theorems which works towards proving not only that every group of order $p_1\cdots p_m$ where $p_i\not\equiv 1\text{ mod }p_j$ for any $i,j\in[m]$ (greatly generalizing the statement that a group of $pq$ for primes $p with $q\not\equiv 1\text{ mod }p$ is cyclic) but also that numbers of this form are the only numbers for which the converse is true (namely every group of order $n$ is cyclic).

September 13, 2011

## The Unit and Automorphism Group of Z/nZ (pt. II)

Point of Post: This is a continuation of this post.

September 10, 2011