# Homework Help: Abstract Alg.-Abelian groups presentation

1. Dec 30, 2009

### TheForumLord

1. The problem statement, all variables and given/known data
Let Cn be a cyclic group of order n.
A. How many sub-groups of order 4 there are in C2xC4... explain.
B. How many sub-groups of order p there are in CpxCpxC(p^2) when p is a prime? explain.
C. Prove that if H is cyclic of order 8 then Aut(H) is a non-cyclic group. WHAT is its order?
D. What is the Automorphism group of an infinite cyclic group?

2. Relevant equations
3. The attempt at a solution
About A-> it's obvious (because it has index 2) that a subgroup of order 4 is normal...But I can't figure out how many subgroups of this form there are...
About the other parts-I've no idea...

I'll be delighted to get guidance about all the parts in this question.

2. Dec 30, 2009

### futurebird

C2XC4

Sine the group is Abelian, all of the subgroups are normal. You have an element of order 4 and an element of order 2.

<a> = {a, 1}
<b> = {b, b^2, b^3, 1}

Then, a basis from the group would be: {b, b^2, b^3, 1, ab, ab^2, ab^3, a}
What other subgroups could we have? Look at what each element generates.

<ab> = {ab, b^2, ab^3, 1}
<b^2> = {b^2, 1}
<b^3> = {b^3, b, 1, b^2} = <b>
<ab^2> = {ab^2, 1}
<ab^3> = {ab^3, b^2, ab, 1} = <ab>

Looks like there are 2 (EDIT TO ADD "cyclic") subgroups of order 4.

Last edited: Dec 30, 2009
3. Dec 30, 2009

### TheForumLord

There is no "brute-force" way to do it?
Anyway, you've just found the cyclic groups of order 4...There might be non-cyclic sub-groups of order 4 in C2xC4...
I'll be glad if you could also help me in the other parts of the question and maybe help me find a non-brute-force way to solve this part...
TNX a lot!

4. Dec 30, 2009

### futurebird

I forgot about the non-cyclic subgroups. Drat. Well the only non-cyclic group of order 4 is the Klien group ... is that a subgroup?

5. Dec 30, 2009

### TheForumLord

Don't think so...

6. Dec 30, 2009

### futurebird

OK. But WHY? I can't think of any reason why it can't be a subgroup... The subgroups of Z2XZ2 are Z2... and they are also subgroups of Z4XZ2... it divides the order of the group... um... Let me think about this. Don't rule it out so fast.

7. Dec 30, 2009

### TheForumLord

Actually there are at least 3 groups of order 4 in this abelian group:
{1}*C4, <(a,b)> [where C2=<a>,C4=<b>] and <(a,b^2) > ... :)
So maybe klein's group does exist in there :)

8. Dec 30, 2009

### futurebird

What is <a, b^2>?

Looks like Z2XZ2 to me. It's a group of order 4 and it's *not* cyclic.

9. Dec 30, 2009

### TheForumLord

So there are 3 ... But is there any not-brute-force way to discover this?

TNX a lot for your help!

10. Dec 30, 2009

### futurebird

Not that I know of. Finding sub groups can be hard. On to B.

"How many sub-groups of order p there are in CpxCpxC(p^2) when p is a prime?"

UM.

We have the three cyclic subgroups. Cp, Cp and C(p^2)

C(p^2) will always have Cp as a subgroup.

So that's three so far. Can there be others? I don't know yet.

This is good practice for me, I'm not an expert, but I always liked this subject.

11. Dec 30, 2009

### TheForumLord

Everything you've said is pretty much trivial... I like this subject too but I'm realy bad at it as you can see... It looks like a pretty simple question but I can't find any elegant way to solve it...Welll...hope you'll be able to help me on this...

BTW- an order p sbgrp is cyclic of course...Hence we need to find all the (a,b,c) in CpxCpxCp^2 such as lcm(o(a),o(b),o(c)) = p... p is a prime so we one of the elements must have order p and the other two order 1 or p... The options are:
p, p, p & p,1,1& 1,p,1& 1,1,p& p,p,1 etc... and these are the only options...
AM I right?

TNX

12. Dec 30, 2009

### futurebird

We are looking for groups of order p, so yes, they will be cyclic. Your notation is confusing me. "Hence we need to find all the (a,b,c) in CpxCpxCp^2 such as lcm(o(a),o(b),o(c)) = p"

Why are we looking at 3 elements at once? Aren't we just looking for the elements with order P?

The elements in CpxCpxCp^2 would be products of:

<a> = {a, a^2, ..., a^(p-1), 1}
<b> = {b, b^2, ..., b^(n-1), 1}
<c> = {c, c^2, ..., c^(pp-1), 1}

The fundamental theorem of finitely generated abelian groups tells us this is an abelian group... right?

Now the elements are (I felt the need to write this out for some reason...):

1 ,a, a^2, ..., a^(p-1), b ,ba, ba^2, ..., ba^(p-1), b^2 ,b^2a, b^2a^2, ..., b^ba^(p-1), b^(p-1) ,b^(p-1)a, b^(p-1)a^2, ..., b^(p-1)a^(p-1)

c ,ca, ca^2, ..., ca^(p-1), cb ,cba, cba^2, ..., cba^(p-1), cb^2 ,cb^2a, cb^2a^2, ..., cb^ba^(p-1), cb^(p-1) ,cb^(p-1)a, cb^(p-1)a^2, ...,cb^(p-1)a^(p-1)
.
.
.
c^(p^2-1) ,c^(p^2-1)a, c^(p^2-1)a^2, ..., c^(p^2-1)a^(p-1), c^(p^2-1)b ,c^(p^2-1)ba, c^(p^2-1)ba^2, ..., c^(p^2-1)ba^(p-1), c^(p^2-1)b^2 ,c^(p^2-1)b^2a, c^(p^2-1)b^2a^2, ..., c^(p^2-1)b^ba^(p-1), c^(p^2-1)b^(p-1) ,c^(p^2-1)b^(p-1)a, c^(p^2-1)b^(p-1)a^2, ...,c^(p^2-1)b^(p-1)a^(p-1)

Horrible, But we can see that in general each one has the form:

(a^x)(b^y)(c^z)

x, y in {1...p}
z in {1...p^2}

keeping in mind: a^p = b^p = c^(p^2) = 1

Which are equal to 1 only if taken to the p power? Maybe this is what you were doing with the lcm? But it's not making sense to me, can you explain more?

Last edited: Dec 30, 2009
13. Dec 30, 2009

### futurebird

We want [(a^x)(b^y)(c^z)]^p = 1

(a^x)^p = 1 only when x=1
(b^y)^p = 1 only when y=1
(c^z)^p = 1 only when z=p

so:

a*1*1
a*b*1
a*b*c^p
1*b*1
1*b*c^p
1*1*c^p

We have 6 possible generators.

Last edited: Dec 30, 2009
14. Dec 30, 2009

### rochfor1

You guys are on the right track for A+B, so I won't say anything about those two.

With respect to C+D, convince yourself that a homomorphism whose domain is a cyclic group is completely determined by where it sends the generator. With this fact in mind, we get the following answers.

C: Any isomorphism preserves the order of group elements (think about why this is), so any automorphism of Z_8 must send 1 to an element of order 8. The elements of order 8 in Z_8 are {1, 3, 5, 7} (because these are coprime to 8), so we see that order of Aut(Z_8) is 4. I'll leave it to you to show that Aut(Z_8) is isomorphic to the Klein four group and is therefore not cyclic.

D: This is essentially the first observation I wrote. Use that fact to show that Aut(Z) is isomorphic to Z.

15. Dec 30, 2009

### Hurkyl

Staff Emeritus
Haven't you done pretty much all of C for them, and the hard part of D? :grumpy: Leave some work for them to do!

Fortunately, they still have to think about the problem -- your advice has at least one significant error, let's see if they can work it out....

16. Dec 30, 2009

### TheForumLord

Well, where should I start... First of all-we do look for 3 elements as we look at the direct product of 3 cyclic groups... About C- It's enough to show that two of the automorphisms has the same order... For example- the aut. that send h to h^3 and h to h^7 both have order 2-so it isn't cyclic...

About D-Aut(Z) isn't isomorphic to Z according to my knowledge...but if we take a cyclic group H that is infinite, it's isomorphic to Z... Z has 2 generators - 1 and -1... Hence there are 2 different automorphisms... The id. one and another one such as:
1-> -1 & -1 -> 1 ... and that's it...

As you can all see, I've managed to solve most of the question all by myself (even before I looked at your guidance) but I rly need your verification... About A-if one of you still has a non-brute-force way to solve it, it will be very helpful :)

TNX a lot to you all!

17. Dec 30, 2009

### rochfor1

Ah yes, you're absolutely right about Z...I was thinking about all homomorphisms Z -> Z...oops

18. Dec 30, 2009

### TheForumLord

And about the other parts? :)

TNX a lot!