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Sep 15 2010 05:51pm
Quote (TheShattered @ Sep 15 2010 11:49pm)
They are indeed worthy


Rawr means ily in dinosaur.
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Sep 15 2010 05:57pm
Quote (TheShattered @ Sep 15 2010 06:42pm)
5 + 1 - 2 ^2





i think *twitch*


FIXED


parenthesis were optional, i decided to change that
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Sep 15 2010 05:57pm
Quote (datdude377 @ Sep 15 2010 11:57pm)
FIXED


parenthesis were optional, i decided to change that


PEMDAS.
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Sep 15 2010 06:12pm
For the purposes of this question, we define a Dihedral Group to be a group D = (x , y) where x and y are distinct involutions

1. Classify all abelian dihedral groups.

2. Let z = xy.

i. Show that zx = zy = z−1 and deduce that H = (z) is a normal subgroup of D.
ii. Prove that xH = yH, and deduce that G | H is cyclic of order 1 or 2.
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Sep 15 2010 06:14pm
Quote (nerobellum @ Sep 15 2010 07:12pm)
For the purposes of this question, we define a Dihedral Group to be a group D = (x , y) where x and y are distinct involutions

1. Classify all abelian dihedral groups.

2. Let z = xy.

i. Show that zx = zy = z−1 and deduce that H = (z) is a normal subgroup of D.
ii. Prove that xH = yH, and deduce that G | H is cyclic of order 1 or 2.


ya ya whatever your wrong, i win.
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Sep 15 2010 06:14pm
Quote (nerobellum @ Sep 15 2010 08:12pm)
For the purposes of this question, we define a Dihedral Group to be a group D = (x , y) where x and y are distinct involutions

1. Classify all abelian dihedral groups.

2. Let z = xy.

i. Show that zx = zy = z−1 and deduce that H = (z) is a normal subgroup of D.
ii. Prove that xH = yH, and deduce that G | H is cyclic of order 1 or 2.


wtfx is that -____o
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Posts: 17,374
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Sep 15 2010 06:14pm
Quote (nerobellum @ Sep 15 2010 04:12pm)
For the purposes of this question, we define a Dihedral Group to be a group D = (x , y) where x and y are distinct involutions

1. Classify all abelian dihedral groups.

2. Let z = xy.

i. Show that zx = zy = z−1 and deduce that H = (z) is a normal subgroup of D.
ii. Prove that xH = yH, and deduce that G | H is cyclic of order 1 or 2.


1. x and y are distinct elements of order 2. Let z = xy which is clearly not an element of {1; x; y}.
Now G = {1; x; y; z} is closed under multiplication and z^2 = 1 so it is also closed under inversion.
Thus G is a group of size 4, and has the following multiplicative structure:
the product of any pair of di erent non-identity elements is the third non-identity element.
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Sep 15 2010 06:15pm
Group Theory, bro.
Don't you know anything?

E: TS can I borrow 5 fg?

This post was edited by nerobellum on Sep 15 2010 06:16pm
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Sep 15 2010 06:16pm
Quote (nerobellum @ Sep 15 2010 08:15pm)
Group Theory, bro.
Don't you know anything?


nope.



lost atm
topic changed to ponies
Discuss.

This post was edited by TheShattered on Sep 15 2010 06:17pm
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Sep 15 2010 06:19pm
Quote (TheShattered @ Sep 15 2010 06:16pm)
nope.



lost atm
topic changed to ponies
Discuss.


I like them there fluffy and fun
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