Non-abelian group and periods of elements

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March 9th 2011, 06:10 AM
Zalren

Non-abelian group and periods of elements

Prove that a non-abelian group $G$ of order 6 must have at least one element $x$ of period 3.

This is what I have so far.

Let $x$ in $G$ be a non-identity element. $x$ can have period 2, 3, or 6.
Period of $x$ cannot be 6 because that make $G$ cyclic, thus abelian.
If $x$ has period 3, we are done.
If $x$ has period 2, ... (stuck here)

Maybe looking at an element $y$ in $G$ and $y$ is not in $<x>$ , but not sure what to do with it.

Thanks in advance for any help.
March 9th 2011, 06:13 AM
Ackbeet

Have you done the Sylow theorems yet? Seems like the Corollary to # 1 there might be useful.
March 9th 2011, 06:20 AM
Ackbeet

Wait, here you go:

I think your $y$ idea is the right idea. Note that $\langle x\rangle\not=G.$ Therefore, there exists $y\in G$ such that $y\not\in\langle x\rangle.$ By hypothesis and assumption, $y\not =e,$ and $\langle y\rangle\not=G$ (or your group is cyclic, hence abelian). But, $|\langle y\rangle|$ divides $|G|,$ because it's a subgroup. Therefore,

$|\langle y\rangle|\in\{2,3\}.$

Now you'll need some clever argument as to why $|\langle y\rangle|\not=2.$ Thinking...
March 9th 2011, 06:33 AM
Zalren

Sorry I have not heard of Sylow Theorems yet.

Is it possible to use cosets somehow? Using the factor group $G/<x>$ and using the coset $y<x>$ ?

I believe the factor group $G/<x>$ has order 3, since we are assuming $x$ has period 2.
March 9th 2011, 08:26 AM
topspin1617

Well, if there are no elements of order 3, then the five non-identity elements must all have order 2 (because as you noted, an element of order 6 would make the group cyclic, hence abelian).

Can you prove that any group whose non-identity elements all have order 2 must be abelian? This would of course give an immediate contradiction to $G$ not being abelian.

EDIT: Sorry, using the word "order" where you use "period".
March 9th 2011, 08:30 AM
Ackbeet

Reply to topspin1617:

That is exactly the lines of thinking I was contemplating. But I'm not sure how to do it.
March 9th 2011, 08:40 AM
Ackbeet

Can we guarantee that $\langle x\rangle\trianglelefteq G?$ If so, then the OP'er's idea in post # 4, at the end, could work, if we can identify $G/\langle x\rangle$ with a subgroup of $G.$ Can we do that?
March 9th 2011, 08:52 AM
topspin1617

Quote:

Originally Posted by Ackbeet

Can we guarantee that $\langle x\rangle\trianglelefteq G?$ If so, then the OP'er's idea in post # 4, at the end, could work, if we can identify $G/\langle x\rangle$ with a subgroup of $G.$ Can we do that?

No, because it's not true: just take an arbitrary subgroup of $S_3$ of order 2, you can show that it isn't normal.

But the fact that a group of exponent 2 is abelian is nearly trivial. Let $G$ be such a group, so that $x^2=e$ , for all $x\in G$ .

Suppose $a,b\in G$ are arbitrary. Consider the relation

$(ab)^2=e$
$abab=e$
$ababb^{-1}=eb^{-1}$
$aba=b^{-1}$
$abaa^{-1}=b^{-1}a^{-1}$
$ab=b^{-1}a^{-1}$ .

But if an element has order 1 or 2, it is clearly its own inverse. So the last relation simplifies to

$ab=ba$

and, since these elements were arbitrary, this implies that $G$ is abelian.
March 9th 2011, 08:56 AM
Ackbeet

Quote:

Originally Posted by topspin1617

No, because it's not true: just take an arbitrary subgroup of $S_3$ of order 2, you can show that it isn't normal.

But the fact that a group of exponent 2 is abelian is nearly trivial. Let $G$ be such a group, so that $x^2=e$ , for all $x\in G$ .

Suppose $a,b\in G$ are arbitrary. Consider the relation

$(ab)^2=e$
$abab=e$
$ababb^{-1}=eb^{-1}$
$aba=b^{-1}$
$abaa^{-1}=b^{-1}a^{-1}$
$ab=b^{-1}a^{-1}$ .

But if an element has order 1 or 2, it is clearly its own inverse. So the last relation simplifies to

$ab=ba$

and, since these elements were arbitrary, this implies that $G$ is abelian.

Ah, very nice. Note to Zalren: this post # 8 pretty much solves your problem, I believe.
March 9th 2011, 11:21 AM
Zalren

So I understand what topspin1617 is saying, but does it apply? I am working under the assumption that $x$ has period 2 and that there exists a $y \in G$ and $y \not\in <x>$ (so we don't know the period of $y$ ). I don't think that the proof topspin1617 gave implies $y$ is commutative, thus not proving that $G$ is abelian and creating a contradiction.
March 9th 2011, 11:23 AM
Ackbeet

What topspin is saying is that suppose there is no element of order 3. Then all five non-identity elements are of order 2. Then you do the computation topspin did, and you show that any two elements commute, contrary to the non-abelian assumption.
March 9th 2011, 11:27 AM
Zalren

Ah I see, thank you!
March 9th 2011, 11:28 AM
Ackbeet

Well, obviously topspin did the majority of the work here, but you're certainly welcome for my contribution.