My Math Forum Subgroups of Abelian Groups

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 June 30th, 2012, 02:06 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 $H= (x \in G : x=y^2 for some y\in G)$, 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 $K= (x \in G: x^2 \in H)$, 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, 08:06 AM #2 Member   Joined: Jun 2012 From: Uzbekistan Posts: 59 Thanks: 0 Re: Subgroups of Abelian Groups The subset $H$ of group $G$ is called subgroup of $G$ if: 1) $a,b\in H\Rightarrow ab\in H$ 2)$a\in H\Rightarrow a^{-1}\in H$. In our case $a,b\in H\Rightarrow a=a_1^2, b=b_1^2$ for some $a_1,b_1\in G$. From here $ab=a_1^2b_1^2=a_1b_1a_1b_1=(a_1b_1)^2$, i.e. $ab\in H$. If $a\in H$ then $a=a_1^2$ for some $a_1\in G$ thus $a^{-1}=(a_1^2)^{-1}=(a_1^{-1})^2$, i.e. $a^{-1}\in H$. Thereby $H$ is subgroup in $G$. Analogically, is proved the second question.
 June 30th, 2012, 09:00 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 $( a_1 ^2)^{-1}=(a_1 ^{-1})^2$
June 30th, 2012, 02:24 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 $( a_1 ^2)^{-1}=(a_1 ^{-1})^2$
Let $(a^2)^{-1}=b$. Then $ba^2=e=a^2b$, where $e-$Identity element of group.
We have:
$(a^{-1})^2=a^{-1}\cdot a^{-1}=a^{-1}\cdot a^{-1}\cdot e=a^{-1}\cdot a^{-1}\cdot (a^2b)=(a^{-1}\cdot a^{-1}\cdot a^2)\cdot b=b$. Therefore $( a_1 ^2)^{-1}=(a_1 ^{-1})^2$. By the way $a^{-1}\cdot a^{-1}\cdot a^2=a^{-1}\cdot (a^{-1}\cdot a)\cdot a= a^{-1}\cdot a=e$. It is not necessary where we put brackets when we have the association law.

 July 2nd, 2012, 09:12 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 $a$ be and element of order 12 in a group $G$: what is the smallest possible positive integer k such that $a^{8k}=e$? and what is the order of $a^8, a^9, a^{10}, a^5$? thank you
 July 2nd, 2012, 02:05 PM #6 Member   Joined: Jun 2012 From: Uzbekistan Posts: 59 Thanks: 0 Re: Subgroups of Abelian Groups $a^{12}=e\Rightarrow a^{12n}=e$ for all $n\in\mathbb N$. Hence $12n=8k\Rightarrow k=\frac{3n}{2}\geq 3$, because $n\geq 2$ (If $n=1$ then $k=\frac{3}{2}\notin\mathbb N$). Thereby the smallest $k$ is $3$ when $n=2$. That's why the order of $a^8$ is $3$. Analogically, $12n=9k\Rightarrow k=\frac{4n}{3}\geq 4$. Hence the order of $a^9$ is $4$...
 July 3rd, 2012, 01:57 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 $ord(a)=m$, and $ord(b)=n$. Prove that the order of (a,b) in $G \times H$ is the lowest common multiple of m and n. -Suppose $(c,d) \in G \times H$ 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 $ord(ab)=mn$ I really appreciate the help.
 July 3rd, 2012, 10:14 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 $G\times H$ as $(a,b)\cdot (c,d)=(ac,bd)$. 1) We have $(a,b)^k=(a^k,b^k)$. Hence $(a,b)^k=e\Leftrightarrow a^k=e_{G},b^k=e_{H}\Leftrightarrow k=ms\; (1)$ and $k=nt\; (2)$ where $s,t\in\mathbb N$. We know that $k$ the smallest number which satisfies the conditions (1) and (2). But the smallest number which satisfies the conditions (1) and (2) is $lcm(m,n)$. Thereby $k=lcm(m,n)$. 2) Since m and n are not relatively prime then $gcd(m,n)=d>1=$. From 1) we know that the order of $(c,d)$ is $lcm(m,n)$. By well known formula $gcd(m,n)\cdot lcm(m,n)=mn$, we find $lcm(m,n)=\frac{mn}{gcd(m,n)}
 July 6th, 2012, 10:04 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 $f$ be a homormorphism of $G$ onto $H$ - For each element $a \in G$ the order of $f(a)$ is a divisor of the order of $a$. -Let m be an integer such that m and $|H|$ are relatively prime. For any $x \in G$ , if $x^m \in ker(f)$ , then $x \in ker(f)$ Thank You
 July 9th, 2012, 09:09 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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