Find the number of Sylow 2-subgroups of the special linear group of order 2 on $\mathbb{Z}$ (modulo 3). I think it will be 1. But I failed to prove it using the counting principle. It has 4 sylow 3-subgroups.

Colten Andrade
2022-09-27
Answered

Find the number of Sylow 2-subgroups of the special linear group of order 2 on $\mathbb{Z}$ (modulo 3). I think it will be 1. But I failed to prove it using the counting principle. It has 4 sylow 3-subgroups.

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Kaya Garza

Answered 2022-09-28
Author has **8** answers

Hints: $G=SL(2,3)$ has 24 elements, hence ${n}_{2}=\mathrm{\#}{\text{Syl}}_{2}(G)=1$ or =3. If ${n}_{2}=1$, then a Sylow 2-subgroup must be normal, which is the case indeed. Show that the Sylow 2-subgroup is isomorphic to the quaternion group $Q$ of order 8. Write down the matrices.

fion74185296322

Answered 2022-09-29
Author has **1** answers

Thanks, good

asked 2022-10-18

A bag contains 8 white marbles and 7 blue marbles. Find the probability of selecting 2 white marbles and 3 blue marbles: (hint: use combinations and fundamental counting principle)

It doesn't say it, but I'm sure there are no replacements. I know it's a combination, but I don't know how 3C2 (enteblack it equals 3) would be the answer because just "3" doesn't sound right.

It doesn't say it, but I'm sure there are no replacements. I know it's a combination, but I don't know how 3C2 (enteblack it equals 3) would be the answer because just "3" doesn't sound right.

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We are supposed to use the Inclusion-Exclusion principle to solve: "how many bridge hands contain exactly 3 clubs, or exactly 5 diamonds or exactly 3 aces?" I know you do something like: |3 clubs|+|3aces|+|5dimonds| - |3clubs AND 3 aces| - |3clubs AND 5 diamonds| -...... +|all 3 intersected| But I'm really struggling with find the cases where the aces may or may not be a club/diamond.

asked 2022-11-15

How many license plates can be made consisting of 2 letters followed by 3 digits (using the fundamental counting principle to solve)? What I know:

1. 26 letters in alphabet, so that means $2\times 26$

2. 10 digits possible (0-9), so that means $3\times 10$

3. FC principle says given m and n options gets you $m\times n$ varieties...

... However, the answer key says "676,000" when I got 1560...

1. 26 letters in alphabet, so that means $2\times 26$

2. 10 digits possible (0-9), so that means $3\times 10$

3. FC principle says given m and n options gets you $m\times n$ varieties...

... However, the answer key says "676,000" when I got 1560...

asked 2022-08-12

In a computer game, when choosing a character there are 5 choices for hair style, 3 choices for eye color, and 7 choices for outfits. How many different characters are possible if you were to choose 1 hair style, 1 eye color, and 1 outfit?

asked 2022-09-06

I tried the Fundamental Counting principle approach: There are 26 black cards to chose from for the first card. That leaves $26\times 25\times 24\times 23$ left for the other four cards. So $\frac{26\times 26\times 25\times 24\times 23}{5!}$ (because we can arrange 5 cards in $5!$ ways). I get 77740. Apparently the answer is: 454480.

I'd appreciate help clearing up my misunderstanding. I'd also like to see the combinations approach. Thanks!

I'd appreciate help clearing up my misunderstanding. I'd also like to see the combinations approach. Thanks!

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License plates are made using 3 letters followed by 2 digits. How many plates can be made if repetition of letters and digits is allowed?

asked 2022-07-07

A group of pre-school children is drawing pictures ( one child is making one picture ) using 12-colours pencil set. Given that

(i) each pupil employed 5 or more different colours to make his drawing; (ii) there was no identical combination of colours in the different drawings; (iii) the same colour appeared in no more than 20 drawings,

find the maximum number of children who have taken part in this drawing activity.

( As each child can be identified with his/her unique combination of colours, the number of children can not exceed

C(5,12) + C(6,12) + C(7,12) +...+ C(12,12)

But how to NARROW it using the condition (iii) ? )

(i) each pupil employed 5 or more different colours to make his drawing; (ii) there was no identical combination of colours in the different drawings; (iii) the same colour appeared in no more than 20 drawings,

find the maximum number of children who have taken part in this drawing activity.

( As each child can be identified with his/her unique combination of colours, the number of children can not exceed

C(5,12) + C(6,12) + C(7,12) +...+ C(12,12)

But how to NARROW it using the condition (iii) ? )