# 群论|MATH10079 Group Theory代写

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Proof. The proof requires a return to the notation. We sketch it, leaving the reader to fill in the details. ${ }^{31}$

For $b \in B$ let $\lambda(b)$ be the smallest integer such that an expansion of $b$ of length $\lambda(b)$ has the form $\left[1_{F}, T\right]$. Let $\nu(b)$ be the largest integer such that a chain $b_{0}b$ where $\nu\left(b^{\prime}\right)<p$; there are only finitely many such $b^{\prime}$. For large $n$, $H^{i}(F, R F) \cong H_{c}^{i}\left(\left|B_{n}\right| ; R\right) \cong \lim {\longrightarrow} H^{i}\left(\left|B{n}\right|,\left|B_{n}\right| \cap|B(p)| ; R\right)$; this follows from

## MATH10079 COURSE NOTES ：

Proof. We need only check that $F \mid X^{n} \times I$ is filtered for given $n$. Let $m$ be such that for every cell $e$ of $X^{n}$ the carrier $C(F(e \times I))$ contains at most $m$ cells. Since $F(e \times I)$ is path connected and $e$ is a cell of this carrier, $F(e \times I) \subset$ $N_{X^{n}}^{m}(C(e))$. Thus for all $i$ we have ${ }^{4} F\left(K_{i} \times I\right) \subset N_{X^{m}}^{m}\left(K_{i}\right)$ and
$$F\left(\left(X^{n} \leq K_{i}\right) \times I\right) \subset N_{X^{n}}^{m}\left(X^{m} \leq K_{i}\right) .$$
Since $(X, \mathcal{K})$ is well-filtered, there exists $j$ such that $N_{X^{n}}^{m}\left(K_{i}\right) \subset K_{j}$. Moreover, given $i$, this $j$ satisfies
$$F\left(\left(X^{n} \subseteq K_{j}\right) \times I\right) \subset N_{X^{n}}^{m}\left(X \subseteq K_{j}\right) \subset X \leq K_{i}$$
The verification of the remaining conditions for a map to be filtered is similar.

# 群论 Group Theory PHYS96019

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$$G=P P \cup P x_{2} P \cup \ldots \cup P x_{p} P,$$
where $x_{1}=e$ has been omitted. Let there be $a_{g}$ left cosets of $P$ in $P x_{j} P$. We have
$$a_{j}=\left[x_{j}^{-1} P x_{j}: x_{j}^{-1} P x_{j} \cap P\right]$$

$$x_{j}^{-1} P x_{j}=\text { order } p=p^{i} \text {. }$$
Hence $a_{j}=1$ or $p t, t \geq 1$. Now $a_{1}=1$ for the double coset $P P=P .$ Also $[G: P]=a_{1}+a_{2}+\ldots+a_{p}$. Now $p \mid[G: P]$, since $[G: P]=p^{m-i} s$, and $i \leqq(m-1)$. Thus the number of $a_{j}$ equal to 1 must be a non-zero multiple of $p$. Now $a_{j}=1$ if and only if
$$x_{j}^{-1} P x_{j}=x_{j}^{-1} P_{x} \cap P \leftrightarrow x_{j}^{-1} P x_{j} \subset P \leftrightarrow x_{j}^{-1} P_{x}=P$$

## PHYS96019 COURSE NOTES ：

Let $H=\langle h\rangle$ and $K=\langle k\rangle$. Then
$$[h, k]=h^{-1}\left(k^{-1} h k\right) \in H,$$
since $H$ is normal. But also
$$[h, k]=\left(h^{-1} k^{-1} h\right) k \in K,$$
since $K$ is normal. Thus
$$[h, k]=h^{-1} k^{-1} h k \in H \cap K={e} .$$
Hence $\quad h k=k h$.
By theorem $2.8 .1$ (5),
$$O(h k)=p q .$$

# 群论  Group Theory MATH32001

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since by ${ }^{*}$ note above $m_{i} n_{i}=|a|$. Thus
$$2(|G|-1)=\sum_{i=1}^{k}\left(|G|-m_{i}\right) .$$

Divide by $|G|$ to get:
$$2-2 /|G|=\sum_{i=1}^{k}\left(1-\frac{1}{n_{i}}\right)$$
(Thus $\left.\sum_{i=1}^{k}\left(1-\frac{1}{n_{i}}\right)<2 .\right)$ We may assume $|G|>1$. Thus
$$1 \leq 2-2 /|G|<2 \text {. }$$
Now each $n_{i} \geq 2$; hence $1 / 2 \leqq 1-\frac{1}{n_{i}}<1$.

## MATH32001 COURSE NOTES ：

Let $x \in A_{1}^{\prime} \cap A_{2}^{\prime}$. Then $x \in A_{1}^{\prime}$ means $x=\left(a_{1}, 0\right)$ and $x \in A_{2}^{\prime}$ means $x=\left(0, a_{2}\right)$. Thus
$$\left(0, a_{2}\right)=\left(a_{1}, 0\right) .$$
This means that $a_{1}=a_{2}=0$. Thus
$$x=(0,0),$$
the zero of $A_{1} \oplus A_{2}$. Hence is satisfied.