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June 30th, 2012, 03:06 AM   #1
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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.
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June 30th, 2012, 09:06 AM   #2
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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.
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June 30th, 2012, 10:00 AM   #3
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Re: Subgroups of Abelian Groups

Thank you for the quick reply. But im confused at some points. But one part confuses me, how does
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June 30th, 2012, 03:24 PM   #4
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Re: Subgroups of Abelian Groups

Quote:
Originally Posted by gaussrelatz
Thank you for the quick reply. But im confused at some points. But one part confuses me, how does
Let . Then , where Identity element of group.
We have:
. Therefore . By the way . It is not necessary where we put brackets when we have the association law.
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July 2nd, 2012, 10:12 AM   #5
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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
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July 2nd, 2012, 03:05 PM   #6
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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 ...
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July 3rd, 2012, 02:57 PM   #7
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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.
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July 3rd, 2012, 11:14 PM   #8
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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.
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July 6th, 2012, 11:04 AM   #9
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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
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July 9th, 2012, 10:09 AM   #10
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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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