All abelian groups of order 3k have a subroup of order 3?

[Boorglar]

I am trying to self-learn group theory from an online pdf textbook, but some of the exercises are tough and the answer section doesn't help much (it only says "True").

Homework Statement

Prove or disprove: Every abelian group of order divisible by 3 contains a subgroup of order 3.

Homework Equations

Not really sure.

The Attempt at a Solution

I think I found a solution, but I am using a theorem which has not yet been proven in the textbook, it was only mentionned and said that they will prove it a few chapters later.

Suppose G is a group of order 3n. The condition that G has a subgroup of order 3 is equivalent to the condition that G has an element of order 3 (because a group of order 3 must be cyclic by Lagrange's Theorem). But this condition is also equivalent to the condition that G has some element of order 3k, where k is 1, 2, ..., or n. (ie a multiple of 3). This is true because if there were such an element g, then g^k would have order 3 because the order of g^k is 3k / gcd( k, 3k ) = 3k/k = 3. And obviously the converse is true.

So I have reduced the problem to that of proving that every abelian group G of order 3n has at least one element whose order is a multiple of 3. And this is where I got stuck. I had to use the Fundamental Theorem of Abelian groups, which is not yet proven.

Assuming this theorem, we know G is isomorphic to the direct product of Z_i where the i are prime powers (including powers of 3 since G has 3n elements). Then consider an element of the form (0,0,...,x,0,...,0) where all entries are 0 except one corresponding to the group Z_3^p. The order of this element must divide 3^p, and is therefore a multiple of 3. By isomorphism, G also has an element of order divisible by 3, and the theorem is proven.Since my proof is assuming a theorem which was not proven yet, I feel it is incomplete. Is there are simpler way to answer the question without using the Fundamental Theorem?

 

[jbunniii]

The fundamental theorem of abelian groups is overkill for this problem. What you are trying to prove is a special case of Cauchy's theorem: if $G$ is a (not necessarily abelian) finite group, and $p$ is a prime factor of $|G|$, then $G$ has an element of order $p$.

A direct proof of the special case where $G$ is abelian and $p=3$ is pretty straightforward by induction. Clearly it's true if $|G| = 3$. If $|G| > 3$, then first show that $G$ must have a nontrivial cyclic subgroup $H$ with $1 < |H| < |G|$. Then 3 must divide either $|H|$ or $|G/H|$. Consider each of these possibilities separately and see if you can conclude that $G$ must have an element of order 3 in either case. The fact that $G$ is abelian ensures that $G/H$ is a group. (Why?)

 

[Boorglar]

Wait, when you say G/H, do you mean the set of all elements in G but not in H? Because in that case G/H clearly cannot be a group, since the identity element is in H.

 

[micromass]

Wait, when you say G/H, do you mean the set of all elements in G but not in H? Because in that case G/H clearly cannot be a group, since the identity element is in H.

He means the quotient group.

 

[Boorglar]

oh well... They didn't talk about quotient groups yet either :P

But I think I get your point. G must have a nontrivial subgroup H since |G| = 3n is composite for n > 1 and so not every element generates G. Then, whatever G/H is, I would guess that |H| * |G/H| = |G| so 3 divides |H| or |G/H|. No matter what, then either H or the group G/H has order 3k for k < n, and by induction it must also have a subgroup of order 3.

 

[jbunniii]

oh well... They didn't talk about quotient groups yet either :P

But I think I get your point. G must have a nontrivial subgroup H since |G| = 3n is composite for n > 1 and so not every element generates G. Then, whatever G/H is, I would guess that |H| * |G/H| = |G| so 3 divides |H| or |G/H|. No matter what, then either H or the group G/H has order 3k for k < n, and by induction it must also have a subgroup of order 3.

So if $H$ has a subgroup of order 3, then you can conclude that $G$ also has a subgroup of order 3. (A subgroup of $H$ is also a subgroup of $G$.) But what about $G/H$? This is not a subgroup of $G$. Unfortunately if you haven't covered quotient groups yet, you won't be able to proceed from here, and you'll have to seek a different proof. Do you know about group actions?

 

[Boorglar]

No. I can tell you what I've seen so far:

Definition of groups
Cyclic groups
Permutation Groups
Cosets and Lagrange's Theorem
Euler's and Fermat's Little Theorem
Isomorphisms and direct products.

 

[jbunniii]

P.S. You're correct that $|H| * |G/H| = |G|$. The fact that you were able to conclude this from the notation means it's a good notation.

If you have learned about cosets, $G/H$ is simply the set of (left) cosets of $H$. This is a group if and only if $H$ is a normal subgroup of $G$, which is automatically true if $G$ is abelian.

 

[jbunniii]

No. I can tell you what I've seen so far:

Definition of groups
Cyclic groups
Permutation Groups
Cosets and Lagrange's Theorem
Euler's and Fermat's Little Theorem
Isomorphisms and direct products.

Hmm, OK. My laptop is about to die, so I'll have to pick this up in a bit. I will have to think about a proof that doesn't involve quotient groups or group actions...

 

[Boorglar]

The problem was in the exercise section of the chapter on Isomorphisms.

Ah yes then if G/H is the set of left cosets of H, then the order of G/H must be [G] and by Lagrange's theorem [G] * |H| = |G|.

What is the binary operation for G/H? Do you multiply cosets like g1H * g2H = (g1g2)H ?

 

[micromass]

The problem was in the exercise section of the chapter on Isomorphisms.

Ah yes then if G/H is the set of left cosets of H, then the order of G/H must be [G] and by Lagrange's theorem [G] * |H| = |G|.

What is the binary operation for G/H? Do you multiply cosets like g1H * g2H = (g1g2)H ?

Yes, that is the operation. But you need to be careful. You need $H$ to be a normal subgroup for this to really make sense. This is no problem here since the group is abelian.