Given $Htriangleleft G, Kle G $, $;Knsubseteq H,;$ and $(G:H) = P$, $;p$ Prime, Prove $HK=G$

By this formula: $$ [G : H] = [G : HK][HK : H] $$Since $[G : H] = p$ is prime, one of $[HK : H]$ and $[H : HK]$ must be $1$. It cannot be $[HK : H]$ because $K

ot subset H$ so $HK

e H$. This forces $[G : HK] = 1$ and hence $G = HK$.

Given $Htriangleleft G, Kle G $, $;Knsubseteq H,;$ and $(G:H) = P$, $;p$ Prime, Prove $HK=G$ 1

1. Exact sequence from $G=G_0supset G_1supset G_2supsetcdotssupset G_r=e$

You misread the exercise. It saysShow how to 'connect' $e$ to $G$ by means of $r$ exact sequences of groups, involving only $e$, $G$ and the quotients $H_i = G_i/G_i1$.So you are not supposed to find a single exact sequence connecting $e$ to $G$, but rather $r$ seperate ones. I think the exercise in that form still contains a typo and should read "involving only $e$, $G_i$ and the quotients $H_i = G_i/G_i1$". Then you can take (writing $1$ for $e$) beginmatrix 1 & to & G_1 & to & G & to & H_0 & to & 1, 1 & to & G_2 & to & G_1 & to & H_1 & to & 1, & & & & vdots & & & & 1 & to & G_r-1 & to & G_r-2 & to & H_r-2 & to & 1, 1 & to & 1 & to & G_r-1 & to & H_r-1 & to & 1. endmatrix

2. A normal subgroup of $G$ is a subgroup of the center of $G$.

Every $p-$group is nilpotent and $Hlhd G$ so $Hcap Z(G)

ot=1$ and since $H$ is of prime order it has to be $Hcap Z(G)=H$

Given $Htriangleleft G, Kle G $, $;Knsubseteq H,;$ and $(G:H) = P$, $;p$ Prime, Prove $HK=G$ 2

3. Prove that GS generates G

By assumption, there is some $g Gsetminus S$. Then $g^-1 Gsetminus S$ as well.And for any $s S$, you also have $sg Gsetminus S$, so just write $s = sgg^-1 (G setminus S) (Gsetminus S)$

4. POLL : G OR N ?

yeh G. fave letter in the alphabet. after P. and C. and H

5. inequality $ [f(G):f(f(G))]leq [G:f(G)] $

My original proof only works for finite group, thus is not good. We may take another way. Suppose for $iin I$, $h_i f(f(G))$ are disjoint left cosets of $f(f(G))$ in $f(G)$. Since $h_iin f(G)$, we may assume $h_i=f(g_i)$. Now we prove $g_i f(G)$ is disjoint left cosets of $f(G)$ in $G$. It suffices to show that $g_i^-1g_j

otin f(G)$ for $i

ot=j$. Suppose on the contrary, if $g_i^-1g_jin f(G)$, we have $f(g_i^-1)f(g_j)in f(f(G))$, i.e., $h_i^-1h_jin f(f(G))$, contradicting the disjointness of $h_i f(f(G))$. Now choose $h_i$ properly so that $h_i f(f(G))$ is a left coset decompostion. We have $[f(G):f(f(G))]=|I|$. But $g_i f(G)$ are disjoint left cosets of $f(G)$ in $G$, so $[G:f(G)]geq |I|$. Thus we are done.

6. Topological Group $G$ totally disconnected $Rightarrow$ $G$ hausdorff?

A space $X$ is Hausdorff if and only if the diagonal $Delta subseteq X times X$ is closed. In the case of a topological group, the diagonal can be described as the preimage of $1$ under the map $Xtimes X to X : (x, y) mapsto xy^-1$, and a point is closed in a disconnected space

7. Give an example of a group $G$ where $n|ord(G)$ but there is no $gin G$ such that $ord(g)=n$

Try $S_3$ with $n=6$, where $S_3$ denotes the symmetric group on $3$ symbols

8. Let $H$ act on $G$ by right multiplication. $displaystyle g_1 in O_g_2 iff g_2^-1 g_1in H$

Since $H$ is acting on $G$ by right multiplication, the orbit of $g_1$ is $g_1H$, the left coset of $H$ by $g_1$. To be more precise $$O_g=hcdot g : hin H= gh^-1 : hin H = gh:hin H= gH. $$ Then the rest of the argument seems fairly clear. If anything is still confusing you, let me know.

9. Recover Poisson bracket on $C[G]$ using the Lie cobracket $delta: g to Lambda^2 g$

You have the Poisson bivector explicitly at every point $exp(a)$ and you know that it is multiplicative. When $delta=dr$ was exact you wrote it like $exp(a).r-r.exp(a)$.You have some functions and you need to decompose their differentials in terms of the basis of one forms that you wrote the bivector in. That way when you write $pi ( df wedge dg)$ you will just be reading off some coefficients. If you knew $f=a_j_1a_j_2cdots$,$g=a_i_1 cdots$ were factorizations where the Poisson bivector was written with a sum of $fracpartialpartial a_k wedge fracpartialpartial a_l$ kinds of terms, you would have written them in a dual pair of bases so the computation would be easier. See Gekhtman Shapiro Vainshtein

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Multi Channel Laser Anti-theft Alarm Circuit
The figure shows the circuit of multi-channel laser anti-theft alarm. VT1, vt2, RL1, RP1, R1, K and vd6 constitute an optical control switch. During the day, the photoresistor RL1 is in a low resistance state under light, VT1 is saturated on, vt2 is off, relay K is not energized, normally open contact J is off, and the alarm circuit does not work; At night, RL1 changes to high resistance state, VT1 is cut off, vt2 is saturated and turned on, K is powered on to make j pull in, the circuit supplies power to the whole machine, the red LED is on, the regulated voltage is 6V, and supplies power to the subsequent circuit after voltage reduction by vd7, vd8 and vd9. Laser rods JB1, RL2, VT3, RP2, R3 and R4 form a laser transmitting and receiving network. When the alarm is in the waiting state, the laser rod emits red laser and irradiates it on the photoresist RL2. RL2 is in a low resistance state, and the base potential of VT3 is very low, which provides trigger current for thyristor vs, vs is turned on, IC2 is powered on, and the horn gives an alarm sound. The alarm can be equipped with up to 10 laser transmitting and receiving networks at the same time. The circuit structure is exactly the same, and C3 is an anti-interference capacitor. An is a reset switch. When the power grid is cut off or the gangster cuts off the power supply, the battery immediately supplies power to the whole machine, ensuring the reliability of the alarm device.Component selection: IC1 is voltage stabilizing integrated circuit 7806, and aluminum heat sink shall be added during use. IC2 is KD-9561 four sound chip, and the circuit is connected to the second sound selection terminal sel2 (fire truck alarm sound). The laser source is a red laser rod with voltage of 4.5V and power
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