Thursday, July 28, 2016

Thoughts and Some Abstract Algebra

This year has been a ridiculously laborious one for me: I enrolled in classes fairly beyond my comfort zone, and these seized the opportunity to savagely rip me to shreds. On top of that, I was cast into the shark-infested waters of tutoring, grading, and reciting: one thing is to be a math Jedi, and another is to successfully convey the subject. I have perennially worked arduously in my life, but let me tell you that I have never worked this hard. I used to brag about studying for twelve to fourteen hours a day every once in a while; this is now my routine. My summer exclusively boiled down to preparing for my prelims, and I still feel nervous about them. The amount of knowledge that I have had to digest in the past two months has been absurd. Paradoxically, it looks like I often spend a torrent of strenuous effort writing long and complicated blog entries, but this is actually my way of relaxing, of taking a break, of quickly letting off stream. I may sound like a  masochist, but I have simply assimilated one of Muhammad Ali's quotes: "I hated every minute of training, but I said, 'Don't quit. Suffer now and live the rest of your life as a champion.'"

Enough of my petty grievances. I will now tackle some abstract algebra questions that I found interesting from my book:

Claim: If $G$ is a finite group and $x\in G$ of odd order $n$, then there is a positive integer $k$ such that $x^{2k}=x$.

Proof: $x^n=e\implies x^nx=x^{n+1}=x^{2k}=x$. $\blacksquare$

Claim: If $G$ is a finite group, $H\leq G$, $N\trianglelefteq G$, and $\gcd\left(\left|H\right|,\left[G:N\right]\right)=1$, then $H\leq N$.

Proof: Let $\varphi:G\to G/N$ be such that $g\mapsto gN$. Then $\varphi$ is a homomorphism, which implies that $\varphi\left(H\right)\leq G/N$, which in turn implies that $\left|\varphi\left(H\right)\right|\mid\left[G:N\right]$. Moreover, $\left.\varphi\right|_H$ is a homomorphism, which implies that $H/\ker\left.\varphi\right|_H\cong\varphi\left(H\right)$, which in turn implies that $\left|\varphi\left(H\right)\right|\mid\left|H\right|$. Therefore, $\left|\varphi\left(H\right)\right|=1$, which implies that $\varphi\left(H\right)=\left\{N\right\}$, which in turn implies that $H\subseteq N$. $\blacksquare$

Theorem (Sylow's): If $G$ is a group of order $p^\alpha m$, where $p$ is prime, $\alpha$ is a positive integer, and $p\not\mid m$, then
  1. there is a subgroup of order $p^\alpha$;
  2. if $P$ is a subgroup of order $p^\alpha$ and $Q$ is a $p$-subgroup, then there is a $g\in G$ such that $Q\leq gPg^{-1}$; and
  3. $n_p\equiv1\pmod p$ and $n_p\mid m$, where $n_p$ is the number of subgroups of order $p^\alpha$.
Claim: A group of order $30=2\cdot3\cdot5$ has a normal subgroup.

Proof: Observe that $n_2\in\left\{1,3,5,15\right\}$, $n_3\in\left\{1,10\right\}$, and $n_5\in\left\{1,6\right\}$. If $n_2=1$, $n_3=1$, or $n_5=1$, then we are done. Therefore, let $n_2=3$, $n_3=10$, and $n_5=6$. However, this implies that there are $3$ non-identity elements of order $2$, $20$ non-identity elements of order $3$, and $24$ non-identity elements of order $5$ for a total of $48$ elements, which is a contradiction. Therefore, the group has a normal subgroup. $\blacksquare$

The following is a list of unique representatives of each isomorphism class of abelian groups of order $216=2^33^3$.

$$\begin{align*}\mathbb Z_{8}\times\mathbb Z_{27}&\cong\mathbb Z_{216}\\\mathbb Z_{8}\times\mathbb Z_{9}\times\mathbb Z_{3}&\cong\mathbb Z_{72}\times\mathbb Z_{3}\\\mathbb Z_{8}\times\mathbb Z_{3}\times\mathbb Z_{3}\times\mathbb Z_{3}&\cong\mathbb Z_{24}\times\mathbb Z_{3}\times\mathbb Z_{3}\\\mathbb Z_{4}\times\mathbb Z_{2}\times\mathbb Z_{27}&\cong\mathbb Z_{108}\times\mathbb Z_{2}\\\mathbb Z_{4}\times\mathbb Z_{2}\times\mathbb Z_{9}\times\mathbb Z_{3}&\cong\mathbb Z_{36}\times\mathbb Z_{6}\\\mathbb Z_{4}\times\mathbb Z_{2}\times\mathbb Z_{3}\times\mathbb Z_{3}\times\mathbb Z_{3}&\cong\mathbb Z_{12}\times\mathbb Z_{6}\times\mathbb Z_{3}\\\mathbb Z_{2}\times\mathbb Z_{2}\times\mathbb Z_{2}\times\mathbb Z_{27}&\cong\mathbb Z_{54}\times\mathbb Z_{2}\times\mathbb Z_{2}\\\mathbb Z_{2}\times\mathbb Z_{2}\times\mathbb Z_{2}\times\mathbb Z_{9}\times\mathbb Z_{3}&\cong\mathbb Z_{18}\times\mathbb Z_{6}\times\mathbb Z_{2}\\\mathbb Z_{2}\times\mathbb Z_{2}\times\mathbb Z_{2}\times\mathbb Z_{3}\times\mathbb Z_{3}\times\mathbb Z_{3}&\cong\mathbb Z_{6}\times\mathbb Z_{6}\times\mathbb Z_{6}\end{align*}$$

Claim: If $p$ is an odd prime and $G$ is a group of order $2p$, then $G$ is a semi-direct product of the form $\mathbb Z_p\rtimes\mathbb Z_2$.

Proof: Let $h\in G$ of order $p$, let $k\in G$ of order $2$, let $H=\langle h\rangle$, and let $K=\langle k\rangle$. Then $H\cap K=\left\{e\right\}$, $G=HK$, and $\left[G:H\right]=2$, which implies that $H$ is normal. Therefore, $G=H\rtimes K\cong\mathbb Z_p\rtimes\mathbb Z_2$. $\blacksquare$

Claim: If $H$ and $K$ are groups and $\varphi:K\to\text{Aut}\left(H\right)$ is a homomorphism, then the identity map between $H\rtimes_{\varphi}K$ and $H\times K$ is an isomorphism if and only if $\varphi$ is the constant homomorphism.

Proof: Let $\left(h_1,k_1\right),\left(h_2,k_2\right)\in H\rtimes_{\varphi}K$ and suppose that the identity map between $H\rtimes_{\varphi}K$ and $H\times K$ is an isomorphism. Then

$$\left(h_1\varphi\left(k_1\right)\left(h_2\right),k_1k_2\right)=\left(h_1,k_1\right)\left(h_2,k_2\right)=\left(h_1h_2,k_1k_2\right),\tag*{$\left(\star\right)$}$$

which implies that $\varphi$ is the constant homomorphism.

Conversely, suppose that $\varphi$ is the constant homomorphism. Then $\left(\star\right)$ holds, which implies that the identity map between $H\rtimes_{\varphi}K$ and $H\times K$ is an isomorphism. $\blacksquare$

Claim: Boolean rings are commutative.

Proof: Let $x$ and $y$ be two elements of a Boolean ring. Then

$$x+y=\left(x+y\right)^2=x+xy+yx+y\implies0=xy+yx\implies xy=yx.\tag*{$\blacksquare$}$$