# Abstract algebra 2 questions

1. Dec 10, 2009

### kewljcs

1. If G is a finite group that does not contain a subgroup isomorphic to Z_p X Z_p for any prime p. prove that G is cyclic

im stumped. i dont understand the 'does not contain a subgroup isomorphoc to Z_p X Z_p part.

ive tried using cauchy's theorem for abelian group: if G is a finite abelian group, and let p be a prime that divides the order of G. then g has an element of order p.

so, Z_p is a finite cyclic gruop, but Z_p X Z_p is not because p, p are not relatively prime.

what else.. :/

2 question: Let G be a group of odd order. prove that the mapping x --> x^2 from G to itself is one to one.

i know that the smallest group of odd order is {I, a, a^-1}. i dont understand the mapping, basically.

2. Dec 10, 2009

### Dick

For the first one you need to think about the classification theorem for finite abelian groups. Basically (Zp)^n for n>2 contains Zp x Zp as a subgroup. For the second one, if your map isn't one to one then there are two elements such that x^2=y^2 and x is not equal to y. Can you use that to show there is an element of order 2? What would that imply?

3. Dec 10, 2009

### kewljcs

i think i got the second one. since the mapping is from G to itself, the order is the same, so we're showing that the kernel of this mapping is 0. right?

so i said that suppose x belongs to the kernel of the mapping, we want to show that x must be the identity, e.
this implies that the mapping is injective. hence, 1 to 1 since we are mapping from G to G.

then, x^2 =e. order of the element divides the order of the group, but |x| =2 => does not divide the order of the grup, which is odd.

does that work?

4. Dec 10, 2009

### kewljcs

im still lost about the first one. we didnt learn about that classification theorem which u mentioned.

5. Dec 10, 2009

### Dick

Actually, I may have spoken too quickly. Are these abelian groups? Somehow I was thinking they are. But your statement really didn't say that explicitly. If they are not I'd better give it some more thought.

Last edited: Dec 10, 2009
6. Dec 10, 2009

### kewljcs

u mean for the 1st question?

also, is my solution to the 2nd question correct?

7. Dec 10, 2009

### Dick

I'm asking if G is abelian for both questions. Your solution for the second one is correct if the mapping x->x^2 is a homomorphism, which it is if G is abelian.

8. Dec 10, 2009

### kewljcs

well, if G is cyclic in the end, then we know that it is abelian.
but im not sure if we can use what we're trying to prove to do that :/

9. Dec 10, 2009

### Dick

You are right on that. If they don't give you that G is abelian, you can't use a fact about the conclusion to prove it. These are both easy questions if you assume G is abelian. If you don't, they start to seem much harder. Got to admit, I'm having some doubts as to whether they are even true. They both smell like abelian group problems. BTW why did you say "cauchy's theorem for abelian groups". You know it's true for any group, right? That may have been what set off my abelian assumptions.

10. Dec 10, 2009

### Hurkyl

Staff Emeritus
A counterexample for the first one is S3.

11. Dec 10, 2009

### Dick

Ok, THANKS Hurkyl! I was really expending all my brain cells on trying to find a counterexample of the second one, and realizing I didn't know that many odd order nonabelian groups. That helps a lot.

12. Dec 11, 2009

### Hurkyl

Staff Emeritus
Oh bleh, neither do I. I spent a few moments trying to make a matrix group, but all the ones I could come up with had even order or were Abelian.

(cheats by looking at a list of all small groups)

Apparently the smallest one is the "Frobenius group of order 21". I don't know what a Frobenius group is, but it's presented by
<a,b | a^3 = b^7 = 1, ba = ab^2>​

13. Dec 11, 2009

### Dick

Nice. Is x->x^2 injective? Just kidding. I really think these questions are about abelian groups anyway. Thanks again for making that obvious.

14. Dec 11, 2009

### kewljcs

so, are you all saying that the first question is wrong?

1. If G is a finite group that does not contain a subgroup isomorphic to Z_p X Z_p for any prime p. prove that G is cyclic

im still lost now. :/

15. Dec 11, 2009

### kewljcs

to paraphrase question 1 :

Prove that a finite Abelian group fails to be cyclic if and only if it has a subgroup
isomorphic to Zp × Zp , for some prime p .

16. Dec 11, 2009

### rasmhop

That's not a paraphrase, but a new statement. You changed the implication to a bi-implication ("only if" to "if and only if") and you added the assumption that G is abelian. For anyone to help you, you really need to take a look at your question and write it down as it was given. As Dick and Hurkyl have deduced your original question 1 was wrong, but if you add the assumption that G is abelian then Dick has already given you some great advice.

If you haven't covered the classification of finitely generated abelian groups yet try the following. Let us prove the contrapositive instead:
If G is not cyclic, then it has a subgroup isomorphic to $\mathbb{Z}_p \times \mathbb{Z}_p$ for some prime p.
Assume G is not cyclic. Let a be an element of G of maximal order. Since G is not cyclic you have $\langle a \rangle \not= G$. Let b be an element in G, but not in the cyclic subgroup generated by a. Let m be the order of a and n be the order of b. Then what is the order of ab and what can we deduce from that? (Remember m was the maximal order of a cyclic subgroup) Can you use this to construct a subgroup of G isomorphic to $\mathbb{Z}_p \times \mathbb{Z}_p$?