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June 30th, 2012, 03:06 AM  #1 
Senior Member Joined: Sep 2011 Posts: 140 Thanks: 0  Subgroups of Abelian Groups
Ok i just started to learn abstract algebra, and im kind of stuck in subgroups. It confuses me when there are questions such as : if , prove that H is a subgroup of given that G is Abelian. or other questions such as : Let H be a subgroup of G and let , prove that K is a subgroup of G. How do you approach these questions? im confused about how i know the definition of the subgroups and its two related theorems and yet i cant prove that they are a subgroup of the bigger group. Can anyone explain to me perhaps by guiding me through those two questions i cant solve above? Thank you for any help to the subject matter. 
June 30th, 2012, 09:06 AM  #2 
Member Joined: Jun 2012 From: Uzbekistan Posts: 59 Thanks: 0  Re: Subgroups of Abelian Groups
The subset of group is called subgroup of if: 1) 2). In our case for some . From here , i.e. . If then for some thus , i.e. . Thereby is subgroup in . Analogically, is proved the second question. 
June 30th, 2012, 10:00 AM  #3 
Senior Member Joined: Sep 2011 Posts: 140 Thanks: 0  Re: Subgroups of Abelian Groups
Thank you for the quick reply. But im confused at some points. But one part confuses me, how does 
June 30th, 2012, 03:24 PM  #4  
Member Joined: Jun 2012 From: Uzbekistan Posts: 59 Thanks: 0  Re: Subgroups of Abelian Groups Quote:
We have: . Therefore . By the way . It is not necessary where we put brackets when we have the association law.  
July 2nd, 2012, 10:12 AM  #5 
Senior Member Joined: Sep 2011 Posts: 140 Thanks: 0  Re: Subgroups of Abelian Groups
ok thank you very much i understood it now. Now i have one more question regarding order of elements, and i dont quite understand this problem:  Let be and element of order 12 in a group : what is the smallest possible positive integer k such that ? and what is the order of ? thank you 
July 2nd, 2012, 03:05 PM  #6 
Member Joined: Jun 2012 From: Uzbekistan Posts: 59 Thanks: 0  Re: Subgroups of Abelian Groups for all . Hence , because (If then ). Thereby the smallest is when . That's why the order of is . Analogically, . Hence the order of is ...

July 3rd, 2012, 02:57 PM  #7 
Senior Member Joined: Sep 2011 Posts: 140 Thanks: 0  Re: Subgroups of Abelian Groups
thank you very much. Im sorry to keep asking questions but i have few more questions regarding cyclic groups and orders  Let , and . Prove that the order of (a,b) in is the lowest common multiple of m and n. Suppose where c has order m and d has order n. Prove that if m and n are not relatively prime, then the order of (c,d) is less than mn. Let a and b commute. If m and n are relatively prime then the I really appreciate the help. 
July 3rd, 2012, 11:14 PM  #8 
Member Joined: Jun 2012 From: Uzbekistan Posts: 59 Thanks: 0  Re: Subgroups of Abelian Groups
As long as you didn't give the definition I define multiple in as . 1) We have . Hence and where . We know that the smallest number which satisfies the conditions (1) and (2). But the smallest number which satisfies the conditions (1) and (2) is . Thereby . 2) Since m and n are not relatively prime then . From 1) we know that the order of is . By well known formula , we find . 3) . Let . Hence for some . Since m and n are relatively prime then . Analogically . From here ... If you are learning the theory of group I recommend you "Fundamentals of abstract algebra" by D.S.Malik, John N. Mordeson, M.K. Sen and "???????? ? ???????" by ?.?. ????????? if you know russian language. 
July 6th, 2012, 11:04 AM  #9 
Senior Member Joined: Sep 2011 Posts: 140 Thanks: 0  Re: Subgroups of Abelian Groups
Thank you very much for your consitent replies. Since the area where i live is pretty limited with maths books, im learning from this book i got called "Abstract algebra by Charles C. Pinter", and so far i think its good, i mean the expnantions are clear, i just hoped that it had more solved problems as examples so i can look at them before i progress to the related section. Do you know this book? Which leads me to my next question: I just started to learn homomorphisms, and im having trouble proving this: Let be a homormorphism of onto  For each element the order of is a divisor of the order of . Let m be an integer such that m and are relatively prime. For any , if , then Thank You 
July 9th, 2012, 10:09 AM  #10 
Senior Member Joined: Mar 2012 Posts: 294 Thanks: 88  Re: Subgroups of Abelian Groups
first of all: since f is a homomorphism: f(a^2) = f(aa) = f(a)f(a) = [f(a)]^2. also, since f(a) = f(ae) = f(a)f(e), it follows that f(e) must be the identity of H, e_H. this first result can be extended by induction on n, to show that f(a^n) = [f(a)]^n, for all natural numbers n (and even for all integers, but we don't need that result). now, suppose ord(a) = a = k. among other things, this means a^k = e. therefore: [f(a)]^k = f(a^k) = f(e), the identity of H. since f(a)^k = e_H, f(a) divides k (it might be smaller, because f, for example, might be the homomorphism that maps EVERYTHING to e_H). ********************** by what we did above, we know that x^m in ker(f) means that f(x^m) = e_H, so [f(x)]^m = e_H, so the order of f(x) divides m. but f(x) is an element of H, so (by lagrange) it divides the order of H, H. hence f(x) divides gcd(m,H) = 1, thus f(x) = 1. but this means that f(x) = e_H, which means that x is in ker(f). 

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