# Abelian Group Question

Im going to post this as one giant question and include my attempt at each part, this was given as a practice midterm question ( our practice midterms are not given solutions) my hope is that i can work threw this with some help and maybe even understand the answer. ( help in getting to and critiques on piece's of my proofs very much appreciated.)

Let $G$ be an abelian group.

(i) Suppose that $a$ is in $G$ and has order $m$ (such that $m$ is finite) and that the positive integer $k$ divides $m$.

Prove that $a^k$ has order $m/k$.

EDIT:

We know that the order of $a^{k}$ is the number of elements in $<a^{k}>$ we also know that $a^{m}=e$ where m is the smallest such positive integer that this is true. now $(a^{k})^{n} = e$ for some n in the integers i claim that the smallest such positive n that yields gives us $a^{kn}=e$ is kn where n is the smallest integer that makes $a^{kn}=e$ thus $a^{kn}=a^{m}$

This leads to kn=m thus the smallest such n that is an integer that does this is n =m/k since $(a^{k})^{m/k} = e$ and it is the smallest such integer. Which leads us to the order of $<a^{k}>$ equals m/k

End

(ii)Let $a$ be in $G$ and $l$ be a positive integer. State the theorem which says $\langle a^l\rangle=\langle a^k\rangle$ for some $k$ which divides $m$.

i have a theorem in my textbook that says as follows.

Let G be a finite cyclic group of order n with $a \in G$ as a generator for any integer m, the subgroup generated by $a^{m}$ is the same as the subgroup generated by $a^{d}$ where d=(m,n) if this is the correct theorem can someone please show me why? thanks

(iii) Suppose that $a$ is in $G$ and has order $m$. Use Part i & ii to show that, for any positive integer $l$, the order $a^l$ divides $m$.

by part 1 $a^{l}$ has order m/l so we have order of m divided by the order of l $(m/(m/l)) = l$ Since l is an integer this division yields an integer with remainder 0 for all l in the integers this makes no sense when u pick l>m

(iv)Suppose that $a$ has order $m$ and $b$ has order $n$ with $\gcd(m,n)$=1.

Prove that $\langle a\rangle \cap \langle b\rangle = \{e\}$.

since m and n are relatively prime this should be easy we know $a^{m} =e=b^{n}$ we know $\langle a\rangle$ is a cyclic subgroup of g and that $\langle b\rangle$ is a cyclic subgroup of g so $\langle a\rangle \union \langle b\rangle$ is a subset of G. since G is abelian this must a cyclic group or its not a subgroup of G so $\langle b\rangle$ must be a subset of $\langle a\rangle$ or $\langle a\rangle$ must be a subset of $\langle b\rangle$ since the element $a^{m-1}$ cannot equal $b^{n-1}$ and both are generations for a and b respectively we have reached a contradiction.

i have no idea what i am doing =)

• I like "defiantly does not equal m" =] – Tara B Mar 15 '13 at 23:12
• Honestly i feel like im looking at this question from the wrong planet somehow... =) – Faust Mar 15 '13 at 23:14
• I changed $<a>$ to $\langle a\rangle$ and gcd$(m,n)$ to $\gcd(m,n)$. – Michael Hardy Mar 15 '13 at 23:40
• @MichaelHardy: Isn't it sufficient to note that in the edit comments? – Tara B Mar 15 '13 at 23:51
• @TaraB : I've tended to assume edit comments are seen only be those who look for them. Am I mistaken about that? – Michael Hardy Mar 15 '13 at 23:55

"We know that the order of $a^k= \langle a^k\rangle" I guess you mean$|\langle a^k\rangle|$(i.e. the order of the subgroup generated by$a^k$)? [Oh, also see my comment about the use of$=$near the end of this critique.] "we also know that$a^m=e$where$m$is the smallest such integer that this is true." Yes, although you should really say 'positive integer' or 'natural number'. From here on you could really do with some more punctuation. It's a little difficult to process. "now$(a^k)^n=e$for some$n$in the integers i claim that the smallest such$n$that yields this is the same as$a^m$" Unlikely, since$a^m$is a group element, not an integer! "the order of the element$a^{kn}$is$kn$where$n$is the smallest integer that makes$a^{kn}=e$" Read this over again and see if that's what you meant to say. (I think you probably meant: "the order of$a^k$is the smallest positive integer$n$such that$a^{kn} = e$.) "thus$kn=m$" I would insert "since$m$is the smallest positive integer such that$a^m=e$". "thus the smallest such$n$[that is an integer] is$n =m/k$since$(a^k)^{m/k}=e$and it is the smallest such integer the order of$\langle a^k\rangle=m/k$." 'that is an integer' is unnecessary, since non-integer powers of group elements aren't even defined. Don't use the symbol$=$in the middle of a sentence instead of the word 'equals' or 'is', _especially if there happens to be maths on either side of it and you end up with what looks like an equation, but it doesn't make sense. Here it looks like you're saying$\langle a^k\rangle=m/k$, which is nonsense, since one side is a group, while the other is a number. Anyway, definitely getting there and miles better than before! Old Answer: OK, I'm just going to focus on the first part, because you already seem quite lost there. What does it actually mean for$g$to have order$n$? (Always a good idea to start with definitions if you're stuck on how to prove something.) It means that$n$is the smallest positive number such that$g^n = 1$. So, for this question, you need to show that if$n=m/k$, then$(a^k)^n = 1$and$(a^k)^{n'}\neq 1$for any positive$n'$less than$n$. You'll want to use the fact that$m$is the smallest positive integer such that$a^m=1$. • Edit NVM i actually knew that before posting the question – Faust Mar 15 '13 at 23:24 • OK. So are you going to have another go? No hurry, although if you can do it reasonably soon I will be able to check it. – Tara B Mar 15 '13 at 23:26 • Im doing it yes we know$a^{m}=e$and that$(a^{k})^{n}=e$so i get$a^{kn}=e$thus the gcd(kn,m)=m where kn is bigger then or exactly m we can still say something about n here, i know that n is less than or equal to m – Faust Mar 15 '13 at 23:30 • How do you know that$(a^k)^n = e$? That needs to be part of your proof. I'm confused by the whole part from 'thus' onwards. Note that$kn = m$by the definition of$n$. – Tara B Mar 15 '13 at 23:32 • Sure, there exists such an integer$n$, but I have defined$n$to be a specific integer, namely$m/k\$. (I just gave it a name to avoid messy superscripts.) – Tara B Mar 15 '13 at 23:35