June 2010

Section C.23 June 2010

Subsection Section I: Groups

Activity C.193. Problem 1.

Let $G$ be a group and $H$ a subgroup of $G\text{.}$ Recall that the centralizer of $H$ in $G$ is

\begin{equation*} C_H(G)=\{g\in G|gh=hg\text{ for all }h\in H\} \end{equation*}

Prove that if $H$ is normal in $G\text{,}$ then so is $C_G(H)$ and that $G/C_G(H)$ is isomorphic to a subgroup of the automorphism group of $H\text{.}$

Solution.

Let $G$ be a group and $H\nsg G\text{.}$ Let $x\in C_H(G)\text{,}$ and consider $gxg\inv\text{.}$ Let $h\in H\text{.}$ As $H\nsg G$ we have $h=gh'g\inv$ for some $h\in H\text{,}$ and thus $g\inv h=h'g\inv$ and $hg=gh'\text{.}$ Consider $gxg\inv h\text{.}$ Since $g\inv h=h'g\inv\text{,}$ we see $gxg\inv h=gxh'g\inv\text{.}$ As $x$ commutes with everything in $H$ we have $gh'xg\inv\text{,}$ and since $hg=gh'$ we have $hgxg\inv\text{.}$ Thus $C_H(G)\nsg G\text{.}$

Let $G$ act on the left cosets of $C_G(H)$ by left multiplication, giving rise to the permutation representation homomorphism $\rho:G/C_G(H)\to\Aut(H)\text{.}$ By the First Isomorphism Theorem we see that $G/C_G(H)$ is isomorphic to a subgroup of the automorphism group of $H\text{.}$

Activity C.194. Problem 2 (*).

Suppose $H$ is a subgroup of a group $G$ and $[G : H] = 7\text{.}$ Prove $G$ contains a normal subgroup $N$ such that $N ⊂ H$ and $[G : N ] \leq 7!$
Prove $7!$ is the best possible bound for the previous part — i.e., prove there is a group $G$ and a subgroup $H$ with $[G : H] = 7$ such that for every normal subgroup $N$ of $G$ with $N ⊂ H\text{,}$ we have $[G : N ] ≥ 7!\text{.}$

Solution.

Let $G$ act on $H$ by multiplication on left cosets. Thus $\rho: G\to\Aut(H)\cong S_7\text{.}$ The kernel of this map is the set of all elements in $G$ such that $\rho(g)=\phi_g$ is the identity, meaning that $\phi_g(xH)=gxH=xH$ for all $xH\in G/H\text{.}$ Then $x\inv gxH=H$ and $x\inv gx\in H\text{.}$
Coming soon!

Activity C.195. Problem 3.

Suppose $G$ is a simple group of order $168 = 2^3 · 3 · 7.$ (Yes, there is such a group.)

How many elements of order $7$ does $G$ have?
Show that $G$ has at least $14$ elements of order $3\text{.}$

Solution.

By Sylow’s Theorems, $|\Syl_7(G)|\equiv 1\mod{7}$ and divides $2^3\cdot 3=24\text{.}$ Thus the only options are $1$ and $8\text{.}$ However, as $G$ is simple there cannot be only one Sylow $7$-Subgroup, as it would be normal. Thus there are $8\text{,}$ each having $7$ unique elements and the identity. Thus there are $49$ elements of order $7\text{.}$
By Sylow’s Theorems, $|\Syl_3(G)|\equiv 1\mod{3}$ and divides $2^3\cdot 7=56\text{.}$ As $G$ is simple there cannot be one, so there must be at least $7\text{,}$ each with $2$ non-identity elements. Thus there must be at least $14$ elements of order $3\text{.}$

Subsection Section II: Rings and Fields

Activity C.196. Problem 4.

Let $A$ be the matrix with entries in $\C$

\begin{equation*} A=\begin{bmatrix} 3 & 0 & 0 & 0\\ 1 & 3 & 0 & 0\\ 1 & 0 & 3 & 0\\ 0 & 2 & 1 & 3\\ \end{bmatrix} \end{equation*}

Find the Jordan canonical form of $A\text{.}$
Is $A$ similar to

\begin{equation*} \begin{bmatrix} 0 & -9 & 0 & 0\\ 1 & 6 & 0 & 0\\ 0 & 0 & 0 & -9\\ 0 & 0 & 1& 6\\ \end{bmatrix}? \end{equation*}

Solution.

First, notice that $A$ is upwards triangular, and thus $\cp_A(x)=(x-3)^4\text{.}$ By the [cross-reference to target(s) "thm-cayley-hamilton-thm" missing or not unique] the minimum polynomial divides the characteristic polynomial, and from the definition of minimum polynomial we know $\mp_A(x)$ is the smallest polynomial such that $\mp_A(A) = 0\text{.}$ Since $\mp_A(x)$ must be a power of $(x-3)^n$ with $n\leq 4\text{,}$ we plug in values of $n$ until we get $0\text{.}$

\begin{equation*} (A-3I)^1=\begin{bmatrix} 0 & 0 & 0 & 0\\ 1 & 0 & 0 & 0\\ 1 & 0 & 0 & 0\\ 0 & 2 & 1 & 0\\ \end{bmatrix} \end{equation*}

Shucks. Moving on,

\begin{equation*} (A-3I)^2=\begin{bmatrix} 0 & 0 & 0 & 0\\ 1 & 0 & 0 & 0\\ 1 & 0 & 0 & 0\\ 0 & 2 & 1 & 0\\ \end{bmatrix} \cdot \begin{bmatrix} 0 & 0 & 0 & 0\\ 1 & 0 & 0 & 0\\ 1 & 0 & 0 & 0\\ 0 & 2 & 1 & 0\\ \end{bmatrix}=\begin{bmatrix} 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0\\ 3 & 0 & 0 & 0\\ \end{bmatrix}, \end{equation*}

which is also not $0\text{.}$ However, multiplying one more time we see $\mp_A(x)=(x-3)^3\text{.}$ By part (2) of the [cross-reference to target(s) "thm-cayley-hamilton-thm" missing or not unique] $(x-3)^3$ is an invariant factor. By part (1) of that same theorem, invariant factors must multiply to $\cp_A(x)\text{,}$ and so the invariant factors are $x-3$ and $(x-3)^3\text{.}$ These are also the elementary divisors. So

\begin{equation*} J_1(3)\oplus J_3(3)=\begin{bmatrix} 3 & 0 & 0 & 0\\ 0 & 3 & 0 & 0\\ 0 & 1 & 3 & 0\\ 0 & 0 & 1 & 3\\ \end{bmatrix}. \end{equation*}
Let

\begin{equation*} B= \begin{bmatrix} 0 & -9 & 0 & 0\\ 1 & 6 & 0 & 0\\ 0 & 0 & 0 & -9\\ 0 & 0 & 1& 6\\ \end{bmatrix} \end{equation*}

And notice that $(B-3I)^2=0\text{.}$ Thus $(x-3)^3$ cannot be the minimal polynomial of $B\text{.}$ Two matrices are only similar if they share the same invariant factors (and thus the same minimum polynomial), so $A$ and $B$ are not similar.

Activity C.197. Problem 5.

Let $R$ be a (not-necessarily commutative) ring let $M$ be a left $R$-module. The annihilator of $M$ in $R$ is defined to be

\begin{equation*} \ann_R(M) = \{ r \in R \mid rm = 0 \text{ for all $m \in M$}\}. \end{equation*}

Prove that $\ann_R(M)$ is a $2$-sided ideal of $R\text{.}$
Suppose $M$ is an abelian group (i.e., a $\Z$-module) such that $|M| = 400$ and $\ann_\Z(M)$ is the ideal generated by $20\text{.}$ How many possibilities, up to isomorphism, are there for $M\text{?}$

Solution.

Let $a,b\in \ann_R(M)\text{.}$ Consider $(a+b)m=am+bm=0\text{.}$ Thus $a+b\in \ann_R(M)\text{.}$ Let $r\in\ann_R(M)$ and consider $-r\text{.}$ Let $m\in M$ and suppose $-rm=n$ for some $n\in R\text{.}$ Add $rm$ to both sides to see that $0=n+rm=n\text{.}$ Thus $n=0$ and $-rm=0$ for all $m\in M\text{.}$ So $-r\in\ann_R(M)\text{.}$

Note that as $0m=0=m0$ for all $m\in M\text{,}$ we know that $0\in\ann_R(M)\text{.}$ As $rm=0$ for all $m\in M\text{,}$ we see that $\ann_R(M)$ is a left sided ideal.

Suppose $mr=n$ for some $m\in R\text{.}$ This time we add $-rm$ to both sides, but as $-r\in\ann_R(M)\text{,}$ we once again find that $mr=0\text{.}$ Notice that this means $rm=mr$ for all $r\in \ann_R(M)$ and $m\in M\text{,}$ and thus that elements of $\ann_R(M)$ commute with elements of $M\text{.}$

Let $a,b\in \ann_R(M)\text{.}$ Consider $m(a+b)=ma+mb\text{.}$ Luckily, we know $ma=am=0=bm=mb,$ and thus that $ma+mb=0+0=0\text{.}$ Hence $a+b\in \ann_R(M)\text{,}$ making $\ann_R(M)$ is a two sided ideal.
By the [cross-reference to target(s) "thm-ftfgab" missing or not unique] and Sunzi’s Remainder Theorem, there are only so many options we have for $M\text{:}$
- $\Z_{16}\times \Z_{25}\text{,}$
- $\Z_{8}\times \Z_{2}\times \Z_{25}\text{,}$
- $\Z_{4}\times \Z_{4}\times \Z_{25}\text{,}$
- $\Z_{4}\times \Z_{2}\times \Z_{2}\times \Z_{25}\text{,}$
- $\Z_{2}\times\Z_{2}\times \Z_{2}\times \Z_{2}\times \Z_{25}\text{,}$
- $\Z_{16}\times \Z_{5}\times \Z_{5}\text{,}$
- $\Z_{8}\times \Z_{2}\times \Z_{5}\times \Z_{5}\text{,}$
- $\displaystyle \Z_{4}\times \Z_{5}\times \Z_{20}$
- $\Z_{4}\times \Z_{2}\times \Z_{2}\times \Z_{5}\times \Z_{5}\text{,}$ and
- $\Z_{2}\times\Z_{2}\times \Z_{5}\times \Z_{20}\text{.}$
However, as ideals are additive subgroups, we know that $M$ needs to contain a cyclic subgroup of order $20\text{.}$ Thus we need only consider decompositions with a $\Z_{20}$ in them, of which there are exactly two:
1. $\Z_{4}\times \Z_{5}\times \Z_{20}$ and
2. $\displaystyle \Z_{2}\times\Z_{2}\times \Z_{5}\times \Z_{20}$

Activity C.198. Problem 6.

Let $F$ be a field and $V$ a vector space (not necessarily finite-dimensional) over $F\text{.}$ Prove that every linearly independent subset of $V$ is contained in a basis for $V\text{.}$

Solution.

Let $I$ be linearly independent in $V\text{.}$ Add elements to it until it spans $V\text{,}$ and denote this new set $S\text{.}$ Let $\cP$ denote the collection of all subsets $X$ of $V$ such that $I\subseteq X \subseteq S$ and $X$ is linearly independent. We make $\cP$ into a poset by the order relation $\subseteq\text{,}$ set containment.

We note that $I \in \cP\text{.}$

Let $\cT$ be any totally ordered subset of $\cP\text{.}$ If $\cT$ is empty, then $\cT$ is (vacuously) bounded above by $I\text{.}$ Assume $\cT$ is non-empty. Let $Z = \bigcup_{Y \in \cT} Y\text{.}$ I claim $Z \in \cP\text{.}$

Given $z_1, \dots, z_m \in Z\text{,}$ for each $i$ we have $z_i \in Y_i$ for some $Y_i \in\cT\text{.}$ Since $\cT$ is totally ordered, one of $Y_1, \dots, Y_m$ contains all the others and hence it contains all the $z_i$’s. Since each $Y_i$ is linearly independent, this shows $z_1, \dots, z_m$ are linearly independent. We have shown that every finite subset of $Z$ is linearly independent, and hence $Z$ is linearly independent. Since $\cT$ is non-empty, $Z \supseteq I\text{.}$ Since each member of $\cT$ is contained in $S\text{,}$ $Z \subseteq S\text{.}$ Thus, $Z \in \cP\text{,}$ and it is clearly an upper bound for $\cT\text{.}$

We may thus apply Zorn’s Lemma to conclude that $\cP$ has at least one maximal element, $B\text{.}$ I claim $B$ is a basis of $V\text{.}$

Note that $B$ is linearly independent and $I \subseteq B \subseteq S$ by construction. We need to show that it spans $V\text{.}$ Suppose not. Since $S$ spans $V\text{,}$ if $S \subseteq \Span(B)\text{,}$ then $\Span(B)$ would have to be all of $V\text{.}$ (For note that if $\Span(S) = V$ and $S \subseteq \Span(B)\text{,}$ then for any $v \in V$ we may write $v = \sum_i c_i s_i$ for $s_i \in S$ and $s_i = \sum_j a_{i,j} b_{i,j}$ with $b_{i,j} \in B$ and hence $v= \sum_{i,j} c_i a_{i,j} b_{i,j}\text{,}$ which implies $\Span(B) = V\text{.}$)

Since we are assuming $\Span(B) \ne V\text{,}$ there must be at least one $v \in S$ such that $v \notin \Span(B)\text{.}$ Set $X := B \cup \{v\}\text{.}$ Notice, $I \subset X \subseteq S$ and, by CITEX , $X$ is linearly independent. This shows that $X$ is an element of $\cP$ that is strictly bigger than $B\text{,}$ contrary to the maximality of $B\text{.}$ So, $B$ must span $V$ and hence it is a basis.

Thus $I$ is contained in the basis $B\text{.}$

Subsection Section III: Linear Algebra and Modules

Activity C.199. Problem 7 (*).

Let $R$ be an integral domain with field of fractions $Q\text{.}$ Let $P$ be a prime ideal of $R$ and let

\begin{equation*} S = \bigg\{ \frac rd\in Q|d\not\in P\bigg\}. \end{equation*}

Show that $S$ is a subring of $Q\text{.}$
Show that

\begin{equation*} I = \bigg\{ \frac pd|p\in P, d\not\in P\bigg\} \end{equation*}

is a prime ideal of $S\text{.}$

Solution.

First, notice that since $P$ is prime, $R\sm P$ is multiplicatively closed. Let $x,y\in S\text{.}$ Thus the denominators of $x$ and $y$ will still not be in $P\text{,}$ making $S$ multiplicatively closed.
Coming soon!

Activity C.200. Problem 8 (*).

Prove $Z[2i] = \{a + 2bi | a \text{ and } b \text{ are integers}\}$ is not a PID.

Hint.

One method is to use (with proof) the fact that $2 + 2i$ is irreducible in this ring.

Solution.

Coming soon!

Activity C.201. Problem 9.

Consider $f(x) = x^6 + 3 \in \Q[x]\text{.}$

Let $a$ be a root of $f(x)$ and prove $\Q(a)$ is a Galois field extension of $\Q\text{.}$
Find the Galois group $\Gal(\Q(a)/\Q)\text{.}$

Hint.

First show $\frac{a^3+1}2$ is primitive $6$-th root of unity.

Solution.

Consider $f(x) = x^6 + 3 \in \Q[x]\text{.}$

Let $a$ be a root of $f(x)\text{.}$ Note that $(\frac{a^3+1}{2})^2=\frac{a^3-1}{2}\text{,}$ and so

\begin{equation*} (\frac{a^3+1}{2})^6=((\frac{a^3+1}{2})^2)^3=(\frac{a^3-1}{2})^3=1, \end{equation*}

so yay! It’s primitive. Using one $6$th primitive root we can obtain all the others, specifically $e^{2\pi i/6}\text{.}$ So we multiply each root by this to get all the others. So we have our splitting field.
Since $\Q(a)$ is Galois, we see that $[\Q(a):\Q]=6\text{.}$ So $\Gal(\Q(a)/\Q)$ is either $S_3$ or $\Z/6.$ The roots of $f$ are $\alpha_i=\sqrt[6]{3}e^{{(\pi i/6)}^{2i-1}}$ for $1\leq i\leq 6\text{.}$

By the Porism there exists a $\sigma\in\Gal(\Q(a)/\Q)$ such that $\sigma(\alpha_1)=\alpha_2\text{.}$ Let $\zeta=e^{\pi i/3}\text{,}$ and note that $\alpha_i=\alpha_{i-1}\zeta\text{.}$ Additionally, note that $\frac{\alpha_i^3+1}{2}=\zeta$ when $i=1,3,5$ and $\frac{\alpha_i^3+1}{2}=\zeta^5$ when $i=2,4,6\text{.}$

Observe then that

\begin{equation*} \sigma(\alpha_2)=\sigma(\alpha_1)\sigma(\zeta)=\alpha_2\frac{\sigma(\alpha_1)^3+1}{2}=\alpha_2\zeta^5=\alpha_1. \end{equation*}

Similarly, we see that $\sigma(\alpha_3)=\alpha_6$ and $\sigma(\alpha_4)=\alpha_5\text{.}$ Thus $\sigma$ corresponds to the permutation $(12)(36)(45)\text{.}$

Using the Porism again we see there exists a $\tau\in\Gal(\Q(a)/\Q)$ such that $\tau(\alpha_2)=\alpha_3\text{.}$ Using a similar process as above we see that $\tau$ corresponds to $(14)(23)(46)\text{.}$ However, observe that $\sigma\tau=(153)(264)\text{,}$ while $\tau\sigma=(135)(246)\text{.}$ Thus these elements do not commute, so we cannot be in $\Z/6\text{.}$ Thus $\Gal(\Q(a)/\Q)\cong S_3\text{.}$

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