Abstract Algebra Thoughts #1
May 4th, 2020

A couple weeks ago, I began my journey (once again) into abstract algebra. I'm hoping that this blog series will depart from the formality of my past solution manuals, and will be more like a discussion, where I can talk abut what I read, things that I realized, and some of the problems I attempted/solved. I am reading Dummit and Foote's textbook on introductory abstract algebra, which comes highly reccomended by a lot of math-people who are much smarter than me. Unfortunately, I don't have much to talk about quite yet, as I am still fairly early in the book. Thus, I'm going to begin by talking about a derivation that I did of one of the things Dummit and Foote assert without a proof, and then I'll jump into my solutions to an assortment of practice problems.

We'll start by discussing the following fact: the Euler totient function, when taking a power of a prime number as an argument is given by:

$$\phi(p^k) \ = \ p^k \ - \ p^{k \ - \ 1} \ = \ p^{k \ - \ 1}(p \ - \ 1)$$
No proof of this fact is provided in the book, but I will provide one here. Consider some $p^k$. Also, consider the definition of the Euler totient function, which is to output the number of coprime integers less than or equal to it's agument (in this case, we want to find the number of coprime integers of $p^k$, such that they are less than or equal to $p^k$). Now, it is fairly easy to see that each integer of the form $ap$ is not comprime with respect to $p^k$. We can also see that the set of all integers of the form $ap$ comprises the set of all integers that are not comprime relative to $p^k$, as $p^k$ is only divisible by some power of $p$, and any integer of the form $ap^n$ can be expressed as $bp$ with $b \ = \ ap^{n \ - \ 1}$. Since the integers that we concerning ousevles with must be less than or equal to $p^k$, it follows that $1 \ \leq \ a \ \leq \ p^{k \ - \ 1}$. Since there are $p^k$ integers from $1$ to $p^k$, we can say that the number of integers that are comprime relative to $p^k$ is given by the diffence of $p^k$ (all possible integers that we can consider) and $p^{k \ - \ 1}$, which is the number of possible values of $a$ that can give us integers that are not comprime to $p^k$. Thus, we arrive at the quantity $p^k \ - \ p^{k \ - \ 1}$.

There's not much else to discuss, so let's venture into the end-of-chapter exercises. A lot of the problems at the beginning of the set seemed either pretty simple, or imply involved rote computation, so I decided to skip to closer to the end. The first problem I attempted was the following: given some integer $n$, and some prime number $p$, find the greatet power of $p$ such that it divides $n! \ = \ n \ \cdot \ (n \ - \ 1) \ \cdot \ (n \ - \ 2) \ ... \ 2 \ \cdot 1$. Esentially, we wish to find the greatest $k$ such that:

$$\frac{n!}{p^k} \ = \ x \ \ \ \ \ \ \ \ \ x \ \in \ \mathbb{Z}$$
I spent an embarassingly long time on this problem. I approached it the complete wrong way from the very beginning. I first thought that we could begin be equating $p^k$ and $n!$, and solve for $k$, and then see where we could go from there, as $1$ i the lowet integer that we could get as the result of this division. Well, this turned out to be a misguided pursuit: equating these two number basically tells us nothing. I then tried to think about $n!$ in term of its prime factorization. We can write $n!$ as:

$$n! \ = \ p_1^{k_1} p_2^{k_2} \ ... \ p_{i}^{k_i}$$
Thus, finding the maximum value of $k$ such that $p^k$ divides $n!$ is the same as finding the power of our given prime $p$ in the prime factorization of $n!$. We can further our argument that realizing, by the definition of the factorial, that the prime factorization of $n!$ will be the product of the prime factorizations of all integers from $1$ to $n$. Now, it is just a matter of identifying the power of a given prime in this product. This is where I got stuck for a while. I played around with a few different ideas for a little while. I wanted to try to derive an expression that would allow me to find the greatest integer $x$ in the "product expansion" of $n!$ such that $p$ is a prime number. As it turns out, this is possible. Consider the expresion $\frac{n}{p}$. If $p$ divides $n$ (meaning, that $p$ is a prime factor of $n$) then this expression will yield an integer. If not, we will get some non-integer real number (an integer $z$, plus some real number $y$ such that $0 \ < \ y \ < \ 1$). Now, consider this integer $z$. Since it is an integer, this could be the result of $\frac{x}{p}$ for some $x \ < \ n$. We can now consider the following proposition:

Proposition: If $n$ is an integer and $p$ is a prime number, then the greatest integer $z$ such that $z \ \leq \ n$ that has $p$ as a prime factor is given by:

$$z \ = \ \left \lfloor{\frac{n}{p}} \right \rfloor p$$
Proof: Let us consider the prime factorization of $z$:

$$z \ = \ p_1^{k_1} p_2^{k_2} \ ... \ p_{i}^{k_i} p^{t} \ = \ \ell p$$
Where $\ell$ is simply the integer that is found from multiplying all of the prime factor of $z$, except for $p$ to the first power. Now, $z$ must be maximized, so we let $\ell$ be the greatest integer less than or equal to $n$ such that $\ell p \ \leq \ n$. We will thus have $\ell \ \leq \ \frac{n}{p}$. Since $\ell$ must be an integer, it must be true that:

$$\ell \ \leq \ \left \lfloor{\frac{n}{p}}\right \rfloor \ \leq \ \frac{n}{p}$$
Maximum $\ell$ occurs at equality, thus:

$$\ell \ = \ \left \lfloor{\frac{n}{p}} \right \rfloor \ \Rightarrow \ z \ = \ \ell p \ = \ \left \lfloor{\frac{n}{p}} \right \rfloor p$$
Which concludes the proof. Now, I originally worked out this fact with little direction, but once I thought about it a bit more, it became more clear that I was heading in the right direction. Consider some integer between from $1$ to $n$ (which is contained in the product expansion of $n!$). if we wish to find the number of numbers in which a given prime factor occurs, we can simply consider the quantity that we just derived. Essentially, we have found the maximum multiple of $p$ that is in the product expansion. to find out how many of these multiples exist, we can simply recognize that any number that is a multiple of $p$ can be written as $ap$, where $a$ ranges from $1$ to $\left \lfloor{\frac{n}{p}} \right \rfloor$ (as, like I said, this is the maximum allowed value by which we can multiply $p$ to get a number in the product expansion of $n!$). Thus, $a$ gives us the number of integersin the product expansion that have $p$ as a factor. We are almost done. The original goal was to find the power to which $p$ is raised in the prime factorization. There is a way that we can modify the tool we have just discovered to find this value: we can evaluate $\left \lfloor{\frac{n}{p^i}} \right \rfloor$ for all powers of $p$, given by $p^i$. The reason this works is because it allows us to "count" the number of powers of a prime in the prime factorization of some given integer from $1$ to $n$. This we, for some prime $p$, we end up "counting it" the same number of times as the power to which it is raised, as such a number will be a multiple of $p^i, \ p^{i \ - \ 1}, \ ..., \ p^{2}, \ p$. If this proposition seemed "hand-wavy", it was. This was simply meant to help build intuition for why we find that the maximum power $k$ such that $p^k$ divides $n!$ is given by:

$$k \ = \ \displaystyle\sum_{p^i \ \leq \ n} \left \lfloor{\frac{n}{p^i}} \right \rfloor$$
for $i \ \ \in \ \mathbb{Z}$. Now, let's do a formal proof using induction. It is easy to check this formula in the case of $n \ = \ 1$. Now, we assume it is true for $n$, and we prove it true for the case of $n\ + \ 1$. We have:

$$\displaystyle\sum_{p^i \ \leq \ n \ + \ 1} \left \lfloor{\frac{n \ + \ 1}{p^i}} \right \rfloor$$
For each term in this sum, there are two cases we must consider, the one where $p^i$ divides $n \ + \ 1$, and the case where it doesn't. If it doesn't, then $\left \lfloor{\frac{n \ + \ 1}{p^i}} \right \rfloor \ = \ \left \lfloor{\frac{n}{p^i}} \right \rfloor$. If it does, then we will have $\left \lfloor{\frac{n \ + \ 1}{p^i}} \right \rfloor \ = \ \left \lfloor{\frac{n}{p^i}} \right \rfloor \ + \ 1$. Thus, we have:

$$k \ = \ \Big( \displaystyle\sum_{p^i \ \leq \ n} \left \lfloor{\frac{n}{p^i}} \right \rfloor \Big) \ + \ s$$
Where $s$ is the number of $i$ such that $p^i$ divides $n \ + \ 1$ (since we add $1$ to the sum each time this happens). At the beginning of the problem, we showed fairly easily that this problem is equivalent to finding the number $k$ such that $k$ is the power of $p$ in the prime factorization of $n!$. We have found that the sum in question is equal to the power of $p$ in the prime factorization of $n!$ (which we will call $k_{n!}$, thus we can write $n! \ = \ bp^{k_{n!}}$, where $b$ does not divide $p$), plus $s$. Let us write $n \ + \ 1 \ = \ ap^{r}$ (where $a$ does not divide $p$), where $r$ is in the natural numbers, plus $0$. For this number, we see that $p^j$ for all $j$ from $1$ to $r$ divide $n \ + \ 1$. Thus, $s$ is the power to which $p$ is raised in the prime factorization of $n \ + \ 1$. Now, we have:

$$(n \ + \ 1)! \ = \ n! \ \cdot \ (n \ + \ 1) \ = \ bp^{k_{n!}} \ \cdot \ ap^{s} \ = \ c p^{k_{n!} \ + \ s}$$
Thus, we see that when we sum together the power of $p$ in $n!$ and $n \ + \ 1$, we get the power of $p$ in $(n \ + \ 1)!$, thus we have proven our proposition for the case of $n \ + \ 1$ and have completed our proof by induction. $\blacksquare$

Now, we can easily test this out. Consider $n \ = \ 10$. Using a computer program, we find that $10! \ = \ 3268800$. Using another computer program, we find the prime factorization of $10!$:

$$10! \ = \ 2^8 \ \times \ 3^4 \ \times \ 5^2 \ \times \ 7$$
Let's try to use our formula to find the $k$ corresponding to $p \ = \ 3$ (which, from the prime factorization, should be $4$). We get:

$$k \ = \ \displaystyle\sum_{3^i \ \leq \ 10} \left \lfloor{\frac{10}{3^i}} \right \rfloor \ = \ \left \lfloor{\frac{10}{3}} \right \rfloor \ + \ \left \lfloor{\frac{10}{9}} \right \rfloor \ = \ 3 \ + \ 1 \ = \ 4$$
So our formula seems to work!

Writing about this solution took up more space than I thought it would, so I'm going to write about one more problem before I wrap this post up. We wish to prove that for some integer $N$, there exist a finite number of integers $n$ such that $\phi(n) \ = \ N$. We then must conclude that as $n \ \rightarrow \ \infty$, then $\phi(n) \ \rightarrow \ \infty$.

We know (as it was written previously in this chapter) that:

$$\phi(n) \ = \ p_1^{\alpha_1 \ - \ 1} (p_1 \ - \ 1) p_2^{\alpha_2 \ - \ 1} (p_2 \ - \ 1) \ ... \ p_s^{\alpha_s \ - \ 1} (p_s \ - \ 1)$$
Where each $p_i$ is a prime factor of $n$. Now, let's assert that $\phi(n) \ = \ N$. Clearly, we must have:

$$p_1^{\alpha_1 \ - \ 1} (p_1 \ - \ 1) p_2^{\alpha_2 \ - \ 1} (p_2 \ - \ 1) \ ... \ p_s^{\alpha_s \ - \ 1} (p_s \ - \ 1) \ = \ N$$
The product of all $p_i^{\alpha_i \ - \ 1}$ will just be some nautral number, so we have:

$$a (p_1 \ - \ 1) (p_2 \ - \ 1) \ ... \ (p_s \ - \ 1) \ = \ N$$
Now, let's absorb all $(p_i \ - \ 1)$ into the natural number $a$, except for one:

$$b (p_k \ - \ 1) \ = \ N$$
Clearly, $b \ \geq \ 1$, so we have:

$$(p_k \ - \ 1) \ \leq \ N \ \Rightarrow \ p_k \ \leq \ N \ + \ 1$$
Thus, for any prime factor in the prime factorization of $n$, if $\phi(n) \ = \ N$, it must be true that any such $p$ is less than or equal to $N \ + \ 1$. Now, let us write the product in a different way. This time, we combine all of the $(p_i \ - \ 1)$, and leave one power of $p_k$:

$$c p_k^{\alpha_k \ - \ 1} \ = \ N \ \Rightarrow \ p_{k}^{\alpha_k \ - \ 1} \ \leq \ N$$
This implies that:

$$\alpha_k \ \leq \ \log_{p_k} N \ + \ 1 \ \leq \ \log_2 N \ + \ 1$$
Since $2$ is the smallest prime number. Let's recap: we have shown that in order for $\phi(n)$ to equal $N$, for some prime factor $p$ in the expansion of $n$, it must be true that $p \ \leq \ N \ + \ 1$. We have also shown that for any prime in the prime factorization of $n$, the power to which that prime can be raised must be less than $\log_2 N \ + \ 1$.

Now, let us pick some $j$. If $\phi(j) \ = \ N$, then $j$ must be equal to the product of all primes less than $N \ + \ 1$, where each prime is raised to some power from $0$ to $\left \lfloor{\log_2 N \ + \ 1} \right \rfloor$. There are a finite number of such combinations of powers to which each prime must be raised, and since each $j$ has a unique prime factorization, it follows that there are a finite number of $j$ such that $\phi(j) \ = \ N$.

From this, we must conclude that $\phi(n) \ \rightarrow \ \infty$ as $n \ \rightarrow \ \infty$. Consider the case where $\phi(n)$ is bounded by some natural number $a$. It would follow that for all $n \ \in \ \mathbb{N}$, that $\phi(n)$ must take on a natural number value from $1$ to $a$. It follows that there must exist at least one natural number $k$ with $1 \ \leq \ k \ \geq \ a$ where $\phi(n) \ = \ k$ for an infinite number of $n$. Thus, $\phi(n)$ cannot be bounded, by what we have already proved. From this, $\phi(n)$ is unbounded, meaning that it must approach $\infty$. Since it is clear that $\phi(n) \ \leq \ n$, $\phi$ can only approach $\infty$ as $n$ approaches $\infty$, thus as $n \ \rightarrow \ \infty$, it follows that $\phi(n) \ \rightarrow \ \infty$.