# Non-abelian group and periods of elements

• Mar 9th 2011, 06:10 AM
Zalren
Non-abelian group and periods of elements
Prove that a non-abelian group $\displaystyle G$ of order 6 must have at least one element $\displaystyle x$ of period 3.

This is what I have so far.

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

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

Thanks in advance for any help.
• Mar 9th 2011, 06:13 AM
Ackbeet
Have you done the Sylow theorems yet? Seems like the Corollary to # 1 there might be useful.
• Mar 9th 2011, 06:20 AM
Ackbeet
Wait, here you go:

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

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

Now you'll need some clever argument as to why $\displaystyle |\langle y\rangle|\not=2.$ Thinking...
• Mar 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 $\displaystyle G/<x>$ and using the coset $\displaystyle y<x>$?

I believe the factor group $\displaystyle G/<x>$ has order 3, since we are assuming $\displaystyle x$ has period 2.
• Mar 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 $\displaystyle G$ not being abelian.

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

That is exactly the lines of thinking I was contemplating. But I'm not sure how to do it.
• Mar 9th 2011, 08:40 AM
Ackbeet
Can we guarantee that $\displaystyle \langle x\rangle\trianglelefteq G?$ If so, then the OP'er's idea in post # 4, at the end, could work, if we can identify $\displaystyle G/\langle x\rangle$ with a subgroup of $\displaystyle G.$ Can we do that?
• Mar 9th 2011, 08:52 AM
topspin1617
Quote:

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

No, because it's not true: just take an arbitrary subgroup of $\displaystyle S_3$ 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 $\displaystyle G$ be such a group, so that $\displaystyle x^2=e$, for all $\displaystyle x\in G$.

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

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

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

$\displaystyle ab=ba$

and, since these elements were arbitrary, this implies that $\displaystyle G$ is abelian.
• Mar 9th 2011, 08:56 AM
Ackbeet
Quote:

Originally Posted by topspin1617
No, because it's not true: just take an arbitrary subgroup of $\displaystyle S_3$ 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 $\displaystyle G$ be such a group, so that $\displaystyle x^2=e$, for all $\displaystyle x\in G$.

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

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

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

$\displaystyle ab=ba$

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

Ah, very nice. Note to Zalren: this post # 8 pretty much solves your problem, I believe.
• Mar 9th 2011, 11:21 AM
Zalren
So I understand what topspin1617 is saying, but does it apply? I am working under the assumption that $\displaystyle x$ has period 2 and that there exists a $\displaystyle y \in G$ and $\displaystyle y \not\in <x>$ (so we don't know the period of $\displaystyle y$). I don't think that the proof topspin1617 gave implies $\displaystyle y$ is commutative, thus not proving that $\displaystyle G$ is abelian and creating a contradiction.
• Mar 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.
• Mar 9th 2011, 11:27 AM
Zalren
Ah I see, thank you!
• Mar 9th 2011, 11:28 AM
Ackbeet
Well, obviously topspin did the majority of the work here, but you're certainly welcome for my contribution.