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parthenos

Member

Registered: 2008-08-25

Posts: 9

abstract algebra - please help

Let H be a subgroup of order 2 in G.  Show that NG(H) = CG(H).  Deduce that if NG(H) = G then H < Z(G).

Last edited by parthenos (2008-09-07 09:23:18)

Ricky

Moderator

Registered: 2005-12-04

Posts: 3,791

Re: abstract algebra - please help

Remember that the difference between normalizer and centralizer is that the normalizer allows you to move around in your subset (group in this case) while the centralizer does not.  How much movement can there really be in a group of order 2?

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"In the real world, this would be a problem.  But in mathematics, we can just define a place where this problem doesn't exist.  So we'll go ahead and do that now..."

parthenos

Member

Registered: 2008-08-25

Posts: 9

Re: abstract algebra - please help

I have no idea truly, my professor doesn't teach us anything.  He just does proofs and give tons of homework.  Please give me more info.  Thanks

Ricky

Moderator

Registered: 2005-12-04

Posts: 3,791

Re: abstract algebra - please help

Do you know the definition of centralizer and normalizer?  With no additional information (other than the very basic properties of a group), that should be enough to solve this problem.

Start with the definition of a normalizer and expand on it case wise for each element in H.

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"In the real world, this would be a problem.  But in mathematics, we can just define a place where this problem doesn't exist.  So we'll go ahead and do that now..."

sa

Guest

Re: abstract algebra - please help

i cant do  algebra

cmcneil

Member

Registered: 2008-09-04

Posts: 3

Re: abstract algebra - please help

I am trying to find the cycle decompositions of some permutations. If you have 2 cyclic groups and you need to find the cycle decomposition of given permutations, I think I got it when there is only one permutations.  You just map the numbers together until cycles form. But, when you have to join 2 groups together like a composition, all I know is let's say you have 2 cyclic groups you have to do permutations with (let's call them a and b). a and b individually will just be the numbers mapping to each other until you come back to the beginning number.  That is one cycle, then you start with the next disjoint smallest number in the group of numbers.

Question:  What if the permutation is a^2, ba, ab, ab^2, or a^b?  I think applying the right side first is how you start, but I am really lost on this one. Please Help!

Ricky

Moderator

Registered: 2005-12-04

Posts: 3,791

Re: abstract algebra - please help

Do you know cycle notation?  In other words, does something like:

(142)(35)

Make sense to you?  Assuming that you do, composing two permutations is fairly easy:

(142)(132)

Now it depends on your notation.  You can either multiply from the right or left.  I will do left:

(142)(132)x

Where x is some number between 1 and 4.  It is typical to start with the lowest number:

(142)(132)1

Now we put the 1 inside the (132) cycle, and we see that 1 gets sent to 3.  Then putting this 3 inside the (142) cycle, we see that 3 gets sent to itself.  So we write down:

(13

Now we want to see where 3 gets sent:

(142)(132)3

Again, put the 3 inside (132), we get that 3 goes to 2.  Now putting 2 inside (142), 2 goes to 1.  So we close of our current cycle:

(13)

Since this means that 3 goes to 1.  Now we start with the next lowest number, 2:

(142)(132)2

Inside of (132), 2 gets sent to 1.  Inside (142), 1 gets sent to 4.  So we write:

(13)(24

Finally, plugging 4 in:

(142)(132)4

(132) fixes 4 (it doesn't move) and inside of (142), 4 gets sent to 2.  So we close off our current cycle:

(13)(24)

And this is our answer.  (142)(132) = (13)(24)

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"In the real world, this would be a problem.  But in mathematics, we can just define a place where this problem doesn't exist.  So we'll go ahead and do that now..."