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\begin{align*}
K &= { \varnothing, { c } } \
&= K + { c } \
K + { a } &= { { a }, { a, c } } \
&= K + { a, c } \
K + { b } &= { { b }, { b, c } } \
&= K + { b, c } \
K + { a, b } &= { { a, b }, { a, b, c } } \
&= K + { a, b, c } \
\end{align*}
Applying the function to the cosets, we get:
\begin{align*}
f(K) &= { \varnothing } \
f(K \cap { a }) &= { { a } } \
f(K \cap { b }) &= { { b } } \
f(K \cap { a, b }) &= { { a, b } } \
\end{align*}
Thus,
$$ f : P_3 \underset{{ \varnothing, { c } }}{\relbar\joinrel\twoheadrightarrow} P_2 $$
$$ P_2 \cong P_3 / { \varnothing, { c } } $$
Q5
$\mathbb{Z}_3$ and $(\mathbb{Z}_3 \times \mathbb{Z}_3) / K$ where $K = { (0, 0), (1, 1), (2, 2) }$
$$ \text{for every } x \in G \qquad \phi_a(x) = axa^{-1} $$
Prove every inner automorphism is an automorphism of $G$.
$$ \phi_a(x) = axa^{-1} $$
Show homomorphic property:
$$ \phi_a(xy) = axya^{-1} $$
But $e = a^{-1} a$, so:
$$ \phi_a(xy) = ax(a^{-1} a)ya^{-1} = \phi_a(x) \phi_a(y) $$
So $\phi_a$ is homomorphic.
Also $\phi_e(x) = x \quad \forall x \in G$
Q3
Likewise from above:
$$ \phi_a \cdot \phi_b = \phi_{ab} $$
Because $a(bxb^{-1})a^{-1} = (ab)x(ab)^{-1}$
For the inverse, we note that:
\begin{align*}
\phi_a(x) \phi_b(x) &= \phi_e(x) = x \
&= (ab) x (ab)^{-1}
\end{align*}
It therefore follows that the inverse automorphism of $\phi_a$ is:
$$ (\phi_a)^{-1} = \phi_{a^{-1}} $$
Q4
$I(G) = { \phi_a : a \in G }$. Prove $I(G) \leq Aut(G)$.
Closure: for any $\phi_a, \phi_b \in I(G)$, then $\phi_a \cdot \phi_b \in I(G)$ because $\phi_a \cdot \phi_b = \phi_{ab}$
Identity: $\phi_e$ is the identity because $eae^{-1} = a$, so $\phi_e \in I(G)$.
Inverses: $\forall \phi_a \in I(G)$, there is an $\phi_{a^{-1}} \in I(G)$ because $\phi_a \cdot \phi_{a^{-1}} = \phi_{a a^{-1}} = \phi_e$, thus $\phi_{a^{-1}} = (\phi_a)^{-1}$
Q5
$$ C = { a \in G : ax = xa \text{ for every } x \in G } $$
Let $a \in C$. Then for every $x \in G$:
$$ ax = xa \text{ or } axa^{-1} = x $$
Q6
Let $h: G \rightarrow I(G)$ be a function defined by $h(a) = \phi$. Prove that $h$ is a homomorphism from $G$onto$I(G)$ and that $C$ is its kernel.
We can see that $h(ab) = \phi_{ab} = \phi_a \cdot \phi_b = h(a) h(b)$. Lastly the function is surjective (onto) because for every $\phi$, there is a corresponding $a \in G$ (possibly multiple if for example the group is abelian), so the mapping is well defined.
The kernel is defined by:
$$ K = { x \in G : f(x) = e } $$
In our case this is:
$$ K = { a \in G : h(a) = \phi_e } $$
The center is defined as:
$$ C = { a \in G : axa^{-1} = x \text{ for every } x \in G } $$
Which is also the same as writing:
$$ K = { a \in G : h(a) = \phi_e } $$
Q7
Lastly using the FHT, we note that:
$$ h : G \underset{C}{\relbar\joinrel\twoheadrightarrow} I(G) $$
$$ I(G) \cong G / C $$
E. FHT Applied to Direct Products of Groups
Q1
Let $G$ and $H$ be groups.
Suppose $J \trianglelefteq G$ and $K \trianglelefteq H$
$$f(x, y) = (Jx, Ky)$$
Assuming $x \in G$ and $y \in H$, then $Jx$ and $Ky$ form the cosets for $G$ and $H$.
That is for every value from $G$ and $H$ maps onto $(G / J) \times (H / K)$ because:
Both $H$ and $K$ are closed subgroups, so an element in both must by definition remain within $H \cap K$.
Let $h \in H \cap K$, then $\forall x \in G, xax^{-1} \in H$. This also applies to $K$. Therefore $H \cap K$ is a normal subgroup of $K$.
Q2
$HK = { xy : x \in H \text{ and } y \in K }$. Prove $HK$ is a subgroup of $G$.
Let $a, b \in HK$, then $ab = (h_1 k_1)(h_2 k_2) = h_1 (k_1 h_2 k_1^{-1}) k_1 k_2$ which is another element in $HK$.
Q3
$H$ is a normal subgroup of $HK$.
Since $HK$ is a subgroup of $G$ then every element of $H$ conjugated with elements from $HK$ also lay within $H$.
$H \trianglelefteq HK$
Q4
Let $x \in HK$ then $x = hk$ for some $h \in H, k \in K$. Form the coset $Hx = H(hk) = Hk$.
Thus $HK/H$ may be written as $Hk$ for some $k \in K$.
Q5
Prove $f(k) = Hk$ is a homomorphism $f: K \rightarrow HK/H$, and its kernel is $H \cap K$
Since $Hk_1 = Hk_2$ for $k_1, k_2$ in the same coset, then any member of the quotient group $HK/H$ is equal to $H$ multiplied by a representative from that member.
To find the kernel, we need every $x \in K$ such that $f(x) = H$, the identity coset. That is $x \in H$. But since we are mapping from $K$, then $x \in K$ and $x \in H$. In other words, $ker f = H \cap K$.
Q6
$$ f : K \underset{H \cap K}{\relbar\joinrel\twoheadrightarrow} HK/H $$
$$ K/(H \cap K) \cong HK/H $$
G. Sharper Cayley Theorem
Q1
To prove $\rho_a$ is a permutation of $X$, we must show it is a bijective mapping from $X$ to $X$.
To show it is injective, let $x_1, x_2 \in X$ and $a \in G$. Suppose $\rho_a(x_1 H) = \rho_a(x_2 H)$. Since $a \in G$ and $G$ is a group, then $a^{-1} \in G$. Then $(a x_1) H = (a x_2) H$ and,
$$ x_1 H = a^{-1} a x_2 H = x_2 H $$
Therefore $\rho_a$ is injective.
To show it is surjective, consider $g \in G$ such that $\rho_a(x) = gH$. But we note that $\rho_a(x) = (ax)H$, so:
$$ gH = axH \text{ or } x H = a^{-1} g H $$
Thus $\rho_a$ is both injective and surjective and is therefore a bijective mapping from $X \rightarrow X$.
Q2
Prove $h : G \rightarrow S_X$ defined by $h(a) = \rho_a$ is a homomorphism.
Let $\rho_e$ denote an identity permutation which leaves the coset unchanged.
$$ \rho_e(xH) = xH $$
$$ h(a) = \rho_a \implies \forall x \in G \qquad \rho_a(xH) = axH = xH $$
But because $\rho_a$ is an identity permutation then $axH = xH$. That is,
$$ xax^{-1} H = H $$
Thus the kernel of $h$ is:
$$ ker f = { a \in H : xax^{-1} \in H, \forall x \in G } $$
Q4
Since $h$ is a homomorphism by:
$$ f : G \underset{ker f}{\relbar\joinrel\twoheadrightarrow} S_x $$
$$ G / ker f \cong \bar{S} \leq S_X $$
If group is a normal subgroup then $\forall a \in A$ and $x \in G$, $xax^{-1} \in A$, which is contained in the kernel of $f$ from the last exercise.
If $H$ contains no normal subgroup of $G$ except ${ e }$ then:
$$ ker f = { e }$$
So the quotient group $G / ker f$ is simply $G$, so we have:
$$ G \cong \bar{S} \leq S_X $$
Since $S_X$ is a permutation representation, for which we only define permutations depending on the elements in $G$. This is why the identity is an homomorphism and not an isomorphism.
\begin{align*}
\cis (x + y) &= (\cis x)(\cis y) \
&= \cos(x + y) + i \sin(x + y) = (\cos x + i \sin x)(\cos y + i \sin y) \
&= \cos(x + y) + i \sin(x + y) = \cis (x + y)
\end{align*}
Q2
$$T = { \cis x : x \in \mathbb{R} } $$
Properties of a group:
Closure
Associativity
Identity
Inverses
Let $u, v \in T$, then the group operation is multiplication and $u = \cis x$ for some $x \in \mathbb{R}$ and $v = \cis y$ for some $y \in \mathbb{R}$.
Then $u \cdot v = (\cis x)(\cis y) = \cis(x + y)$, where $x + y \in \mathbb{R}$ and so $u \cdot v \in T$ which obeys closure property.
Since the result of $\cis$ is a complex number, we conclude the group obeys associativity property.
For the identity, we must test whether 1 lies in $T$. That is $\exists x \in R: \cis x = 1 = \cos x + i \sin x$. Setting $x = 0$, we get $\cis x = 1$, so group obeys identity property.
$$ f : G \underset{K}{\relbar\joinrel\twoheadrightarrow} H $$
$$ S \leq H $$
$$ S^* = { x \in G : f(x) \in S } $$
Q1
Prove $S^* \leq G$
Let $x, y \in S^*$, then $f(x) \in S$ and $f(y) \in S$
Since $f$ is a homomorphism then $f(xy) = f(x)f(y) \in S$
So $xy \in S^*$
Q2
Prove $K \subseteq S^*$
$$ K = { x \in G : f(x) = e_H } $$
$e_H \in S$ because $S$ is a group.
Thus $K \subseteq S^*$
Q3
Let $g$ be the restriction of $f$ to $S^$. That is, $g(x) = f(x)$ for every $x \in S^$ and $S^$ is the domain of $g$. Prove $g$ is a homomorphism from $S^$ onto $S$ and $K = \ker g$.
$$S \leq H$$
Let $s \in S$, then $g(x) = S$, but definition of $S^* = { x : f(x) \in S }$, thus $x \in S^$ and $g$ is a homomorphism from $S^$ onto $S$.
$K = \ker g$ because $K \subseteq S^*$ and $g(x) = f(x)$
Q4
$$ g : S^* \underset{K}{\relbar\joinrel\twoheadrightarrow} S $$
$|G| = k$ and $p$ is a prime divisor. Assume $G$ is not abelian. Let $C$ be the center of $G$ and $C_a$ be the centralizer of $a$ for each $a \in G$.
Let $k = c + k_s + \dots + k_t$ be the class equation.
Show $G$ has at least one element of order $p$.
Q1
Prove: if $p$ is a factor of $|C_a|$ for any $a \in G$ where $a \notin C$, we are done.
$$C_a = { x \in G : xa = ax } $$
Since $C_a$ is subgroup, then this implies there is an element of order $p$ inside $C_a$ by Lagrange's theorem.
Q2
Prove that for any $a \notin C$ in $G$, if $p$ is not a factor of $|C_a|$ then $p$ is a factor of $(G: C_a)$.
From orbit-stabilizer theorem, orbits are conjugacy classes and stabilizers are centralizers, considering the group acting on itself through conjugation.
$$ O(u) = { g(u) : g \in G } $$
$$ G_u = { g \in G : g(u) = u } $$
$$ C_a = { x \in G : xax^{-1} = a } $$
$$ [a] = { xax^{-1} : x \in G } $$
Let the group action $g(u)$ be conjugation $gug^{-1}$ then $C_a$ is equivalent to $G_u$, and $O(u)$ equivalent to conjugacy class $[a]$. Thus,
$$ (G: C_a) = \frac{|G|}{|C_a|} = |[a]| $$
Since $p$ divides $G$ but not $C_a$, then $p$ divides $(G:C_a)$.
Q3
As shown above, the size of the conjugacy class $[a]$ is $(G:C_a)$
$$ k_i = \frac{|G|}{|C_a|} $$
Where $|G|$ has a prime divisor $p$.
But $k = c + k_s + \dots + k_t$ where $k$ and all $k_i$ are factors of $p$, so $c$ is a factor of $p$.
L. Subgroups of p-Groups (Prelude to Sylow)
A p-group is any group whose order is a power of $p$.
If $|G| = p^k$ then $G$ has a normal subgroup of order $p^m$ for every $m$ between 1 and $k$.
Q1
Prove there is an element in $C$ such that $\ord(a) = p$
$$|G| = p^k \implies |C| \text{ is a multiple of } p$$
Thus there is an $a \in C$ such that $\ord(a) = p$
Let $x \in C$ st $\langle x \rangle = C$, then $x^{tp} = e$ and then $a = x^t$
Q2
Prove $\langle a \rangle$ is a normal subgroup of $G$.
Definition of normal subgroup:
$$\forall a \in H, \forall x \in G, xax^{-1} \in H$$
The center is a normal subgrop.
$\langle a \rangle \subseteq C$, thus $\langle a \rangle$ is a nromal subgroup of $G$
Q3
Explain why it may be assumed that $G/\langle a \rangle$ has a normal subgroup of order $p^{m-1}$
$$|G| = p^k \qquad |\langle a \rangle| = p$$
$$\ord(G/\langle a \rangle) = p^{k - 1}$$
Thus for $m$ from 1 to $k$, there is a normal quotient subgroup of order $p^{m -1 }$.
Use J4 to prove that $G$ has a normal subgroup of order $p^m$.
Correspondence theorem:
$$ f : G \underset{K}{\relbar\joinrel\twoheadrightarrow} H $$
$$ S^* = { x \in G : f(x) \in S } $$
$$ S \cong S^*/K $$
Use the natural homomorphism $f : G \rightarrow G/ \langle a \rangle$ with kernel $\langle a \rangle$
Let $S$ be a the normal subgroup of $G/\langle a \rangle$ whose order is $p^{m - 1}$
Show $S^*$ is a normal subgroup of $G$ and its order is $p^m$
Since the order of $\langle a \rangle$ is $p$, and the order of $S$ is $p^{m - 1}$ then the order of $S^*$ is $p^m$
Both $S$ and $K$ are normal subgroups, thus $S^*$ is normal.
M. p-Sylow Subgroups
Q1
Cauchy's theorem states: If $G$ is a group and $p$ is any prime divisor of $|G|$, then $G$ has at least one element of order $p$.
If $q$ is a prime that divides $|G|$ then there would be an element of order $q$. Thus the order of any p-group is a power of $p$.
Q2
Prove every conjugate of a p-Sylow subgroup of $G$ is a p-Sylow subgroup of $G$.
$gHg^{-1}$ is an inner automorphism hence $|H| = |gHg^{-1}|$
Q3
Let $a \in N$ and suppose the order of $Ka$ in $N/K$ is a power of $p$. Let $S = \langle Ka \rangle$ be the cyclic subgroup of $N/K$ generated by $Ka$. Prove that $N$ has a subgroup $S^$ such that $S^/K$ is a p-group.
$$N = N(K) = { g \in G : gK = Kg }$$
$$f:N \rightarrow N/K$$$$ f(a) = Ka$$
Let $x, y \in S^*$ then $f(xy) = f(x)f(y) \in S$
Hence $xy \in S^$ and $S^ \leq N$. By J4:
$$ S \cong S^* / K$$
$|S|$ is a power of $p$.
$$|S^/K| = (S^ : K) = \frac{|S^*|}{|K|} = |S|$$
Q4
Prove that $S^$ is a p-subgroup of $G$, then explain why $S^ = K$ and why it follows that $Ka = K$.
$$S = \langle Ka \rangle$$$$S^* = { x \in N : Kx \in S }$$
$$S^* \leq N \text{ and } a \in N $$$K \leq N$ because normalizer contains the group itself
Let $x \in K$, then $Kx = K \in S$ thus $x \in S^$, so $K \leq S^$ but $K$ is maximal, hence $S^* = K$ and it follows $Ka = K$.
Q5
$$S \cong S^* / K$$
Hence $S = { K }$
Any $Ka \in N/K$ with order $p$ is equivalent to $K$ the identity.
Q6
$$\ord(a) = p^k \implies a^{p^k} = e$$$Ka^{p^k} = K$, thus order of $Ka$ in $N/K$ is a power of $p$.
If $\ord(a)$ is a power of $p$ then $a \in K$
Q7
If $aKa^{-1} = K$ then $a \in N$
$\ord(a)$ is a power of $p$ then $a \in K$
N. Sylow's Theorem
Let $G$ be a finite group and $K$ a $p$-Sylow subgroup of $G$.
Let $X$ be the set of all the conjugates of $K$.
If $C_1, C_2 \in X$, let $C_1 \sim C_2$ iff $C_1 = aC_2 a^{-1}$ for some $a \in K$
Q1
Prove $\sim$ is an equivalence relation on $X$.
$$X = { aKa^{-1}, \forall a \in G$$
$$C_1, C_2 \in X$$$$C_1 \sim C_2 \text{ iff } C_1 = a C_2 a^{-1} \text{ for an } a \in K$$
For each $C \in X$, prove the number of elements in $[C]$ is a divisor of $|K|$.
Conclude that for each $C \in X$, the number of elemenbts in $[C]$ is either 1 or a power of $p$.
From orbit-stablizer:
$$O(C) = { aCa^{-1} : a \in K } = [C]$$$$G_C = { a \in K : aCa^{-1} = C } = N(C) = N$$
$$|[C]| = (K : N)$$
Let $\phi : N^* \rightarrow [C]$ by $\phi(Na) = aCa^{-1}$
Thus $|O(C)| = |[C]| = \frac{|K|}{|N|}$ and the number of elements in $[C]$ is either 1 or a power of $p$.
Alternative: from M2, every conjugate of $K$ is also a $p$-Sylow subgroup of $G$. Hence from Chapter 14 I10, number of elements in $X_C = [C]$ is a divisor of $|K|$.
Q3
Prove the only class with a single element is $[K]$ (using exercise M7).
\begin{align*}
[K] &= { aKa^{-1} : a \in K } \
&= { K }
\end{align*}
If $|[C]| = 1$ then $C = aCa^{-1} \quad \forall a \in K$ which means $C = K$.
Q4
Prove the number of elements in $X$ is $kp + 1$ usings parts 2 and 3.
$$X = {K, C_2, C_3, \dots }$$
$$X = \bigcup_i [C_i]$$
Where $[C_i] \cap [C_j] = \varnothing$ or $[C_i] = [C_j]$
But $|[K]| = 1$ while all other $C_i$ is a positive power of $p$.
Thus $|X| = 1 + kp$
Q5
Prove that $(G : N)$ is not a multiple of $p$.
$(G:N)$ is the number of equivalency classes that partition $G$, which divides $kp + 1$ (number of elements in $X$). It does not divide $p$, hence $(G:N)$ is a not a multiple of $p$.
Q6
Prove that $(N:K)$ is not a multiple of $p$.
$(N:K) = \frac{|N|}{|K|}$ but $K$ is a $p$-Sylow subgroup so $(N:K)$ is not a multiple of $p$.
$$(G:K) = (G:N)(N:K)$$
We know $(G:K)$ is not a factor of $p$, because $p$ is a factor of $|K|$ (from K2), and M5 states no element of $N/K$ has order a power of $p$.
$\therefore (N:K)$ is not a multiple of $p$.
Q7
Prove $(G:K)$ is not a multiple of $p$.
$$(G:K) = (G:N)(N:K)$$
Q8
Let $G$ be a finite group of order $p^k m$ where $p$ is not a factor of $m$. Conclude every $p$-Sylow subgroup $K$ of $G$ has order $p^k$
The only class with a single element is $[K]$ since $aKa^{-1} = K$, all elements where the order is a power of $p$ are in $K$.
P. Decomposition of a Finite Abelian Group into p-Groups
Let $G$ be an abelian group of order $p^k m$ where $p^k$ and $m$ are relatively prime.
Let $G_{p^k}$ be the subgroup of $G$ consisting of all elements whose order divides $p^k$.
Let $G_m$ be the subgroup of $G$ consisting of all elements whose order divides $m$.
Q1
Prove $\forall x \in G$ and integers $s$ and $t$, $x^{sp^k} \in G_m$ and $x^{tm} \in G_{p^k}$.
$p^k$ and $m$ are coprime. Thus $sp^k + tm = gcd(p^k, m) = 1$
$G_{p^k}$ and $G_m$ are subgroups of order $p^k$ and $m$ respectively because $|G| = p^k m$
From Chapter 14, H: if $H$ and $K$ are normal subgroups of $G$, such that $H \cap K = { e }$ and $G = HK$, then $G \cong H \times K$
Firstly all subgroups of $G$ are normal since the group is abelian.
Lastly we have to prove that $\langle a \rangle \cap G' = { e }$
By Lagrange's theorem $\langle a \rangle \cap G' \leq \langle a \rangle$ and also $G'$.
Thus $|\langle a \rangle \cap G'|$ divides $|\langle a \rangle|$ and $|G'| \implies |\langle a \rangle \cap G'|$ divides $\gcd(|\langle a \rangle|, |G'|) = 1$
$$\therefore |\langle a \rangle \cap G'| = 1 = { e }$$
Q3
Explain why we may assume that $G/H = [Hb_1, \dots, Hb_n]$ for some $b_1, \dots, b_n \in G$
Page 149 Theorem 4 from Quotient Groups: "$G/H$ is a homomorphic image of $G$"
$$f: G \rightarrow G/H$$$$f(x) = Hx$$
Let $x \in G$, then $x = a^{k_0} b_1^{k_1} \cdots b_n^{k_n}$ for some $a, b_1, \dots, b_n \in G$
Since $\ord(Hx) = 1$, this means $\ord(Hb_i^{l_i}) = 1$ and because $\ord(b_i) = \ord(Hb_i)$, thus $\ord(b_i^{l_i}) = 1 \implies b_i = e$.
Lastly $a^{l_0} \cdot e = e \implies a = e$
Q6
If $|G|$ has the following factorization into primes: $|G| = p_1^{k_1} \cdots p_n^{k_n}$, then $G \cong G_1 \times \cdots \times G_n \cong \langle a_1 \rangle \times \cdots \times \langle a_n \rangle$.
As shown in previous exercise, the order of $G$ is the product of the order of each generator for the subgroups.
Lastly chapter 10, E3 showed that is $m$ and $n$ are relatively prime, then the products $a^i b^j (0 \leq i \leq m, 0 \leq j \leq n)$ are all distinct. Thus the products of $a$ and $b$ can be decomposed as unique factors.