# G is a group all of whose elements have order two: a^2=1, prove that it is abelian.

• May 18th 2011, 09:31 AM
imagenius
G is a group all of whose elements have order two: a^2=1, prove that it is abelian.
Please can you help me with this question?

Let G be a group all of whose elements have order two: a^2=1 for every aEG.

(i) Prove that G is abelian

(ii) If G is finite, prove its order is of power of two.

I know that abelian is another way of saying commutative, which I understand this as being when you apply a group operation to two group elements, that the result does not depend on its order.
However, I do not understand how to apply this to my question, and do not understand where a^2=1 will come into it (unless obviously its for part (ii)).

I don't know where to begin on part(ii) and are unsure if part (i) will help me with this.
Any help will be great
Thanks
• May 18th 2011, 10:25 AM
Swlabr
Quote:

Originally Posted by imagenius
Please can you help me with this question?

Let G be a group all of whose elements have order two: a^2=1 for every aEG.

(i) Prove that G is abelian

(ii) If G is finite, prove its order is of power of two.

I know that abelian is another way of saying commutative, which I understand this as being when you apply a group operation to two group elements, that the result does not depend on its order.
However, I do not understand how to apply this to my question, and do not understand where a^2=1 will come into it (unless obviously its for part (ii)).

I don't know where to begin on part(ii) and are unsure if part (i) will help me with this.
Any help will be great
Thanks

let a and b be any two elements in your group. then $(ab)^2=1$. As $a=a^{-1}$ and $b=b^{-1}$ then...
• May 18th 2011, 10:25 AM
abhishekkgp
Quote:

Originally Posted by imagenius
Please can you help me with this question?

Let G be a group all of whose elements have order two: a^2=1 for every aEG.

(i) Prove that G is abelian

(ii) If G is finite, prove its order is of power of two.

I know that abelian is another way of saying commutative, which I understand this as being when you apply a group operation to two group elements, that the result does not depend on its order.
However, I do not understand how to apply this to my question, and do not understand where a^2=1 will come into it (unless obviously its for part (ii)).

I don't know where to begin on part(ii) and are unsure if part (i) will help me with this.
Any help will be great
Thanks

assume there are more than two elements in G(if there are only two elements then there is nothing to prove).
let $x,y$ be two distinct elements in G. then $xy \in G$. hence $(xy)^2=xyxy=1 \Rightarrow xyx=y \Rightarrow xy=yx$. (can you justify this?)

for the second part apply cauchy's theorem. the fact that $a^2=1$ for all $a \in G$ will be required.
• May 18th 2011, 10:35 AM
imagenius
thanks for your reply. I understand where the 1st thing is coming from but sorry I dont understand the 2nd part you've found what n equals?
• May 18th 2011, 10:41 AM
abhishekkgp
Quote:

Originally Posted by imagenius
thanks for your reply. I understand where the 1st thing is coming from but sorry I dont understand the 2nd part you've found what n equals?

i don't understand what exactly are you asking.... can you please repost?
• May 18th 2011, 10:47 AM
imagenius
Quote:

Originally Posted by abhishekkgp
assume there are more than two elements in G(if there are only two elements then there is nothing to prove).
let $x,y$ be two distinct elements in G. then $xy \in G$. hence $(xy)^2=xyxy=1 \Rightarrow xyx=y \Rightarrow xy=yx$. (can you justify this?)

for the second part apply cauchy's theorem. the fact that $a^2=1$ for all $a \in G$ will be required.

Ok I understand the 1st bit about how to prove its abelian now. So am I right in saying a^2=1 is used for (ii)?
• May 18th 2011, 11:20 AM
abhishekkgp
Quote:

Originally Posted by imagenius
Ok I understand the 1st bit about how to prove its abelian now. So am I right in saying a^2=1 is used for (ii)?

we had taken two arbitrary distinct elements in G and we showed that they commute. this is what it takes G to be abelian. that's settled.
in the second part we do require that a^2=1 for all a in G. see cauchy's theorem to prove the second part.
• May 19th 2011, 02:17 AM
Deveno
suppose q is an odd prime dividing |G|. then (by cauchy's theorem), G has an element of order q. but all the elements of G satisfiy a^2 = e.

so...?
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