Help. (groups)

LuisRodg · 2009-03-25 05:22:00

The solution to this problem is in the Solutions Manual. Here it is:

We havent covered fields in class, nor Zorn's lemma. I wonder why they put this kind of solution when it uses stuff we havent covered. Is there a way to do this problem using only groups, like, not using any of the stuff that the solution says "the reader is not yet prepared for" ?

Last edited by LuisRodg (2009-03-25 05:22:24)

Ricky · 2009-03-25 06:22:34

If you ignore the infinite case, there is no need for Zorn's Lemma. So we are assuming G is finite. You should be ok with the fact that G is abelian and x = -x. Do you know the fundamental theorem for finite(ly generated) abelian groups? That every finite abelian group is isomorphic to a finite number of copies of the integers modulo p (i.e. cyclic groups). Assuming you know this, consider the map that just moves around these generators. It is an automorphism and you can choose it to be non-trivial as long as G has more than one copy of the integers modulo p. We conclude from this that G is in fact the integers modulo p. But the requirement of x = -x now forces p = 2.

Daniel123 · 2009-03-25 06:59:02

"but the poor reader is not prepared for this"

what a strange book.

daonly1pinkcutie · 2009-03-25 09:50:02

hey could you help me with completing squares? its something that we are doing in algebra 1 right now

LuisRodg · 2009-03-25 09:52:30

Please make a new post and we will try to help you.

daonly1pinkcutie · 2009-03-25 09:59:47

hey wats up my name is brittney could you help me with my math homework?::D

Math Is Fun Forum

#1 2009-03-25 05:22:00

Help. (groups)

#2 2009-03-25 06:22:34

Re: Help. (groups)

#3 2009-03-25 06:59:02

Re: Help. (groups)

#4 2009-03-25 09:50:02

Re: Help. (groups)

#5 2009-03-25 09:52:30

Re: Help. (groups)

#6 2009-03-25 09:59:47

Re: Help. (groups)

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