# fun with "necessarily" abelian

• Oct 25th 2010, 02:35 PM
MichaelMath
fun with "necessarily" abelian
I have this problem:

Suppose that $\displaystyle X$ is a group with $\displaystyle x^2=1$ for all $\displaystyle x \in X$. Show that $\displaystyle X$ is necessarily abelian.

Prove that if $\displaystyle X$ is finite, then $\displaystyle |X|=2^n$ for some $\displaystyle n \geq 0$ and $\displaystyle X$ needs at least $\displaystyle n$ generators.

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For the first part I did:

$\displaystyle \forall x \in X$ , $\displaystyle x(1)=x=(1)x$, so: $\displaystyle e_X=1$

and: $\displaystyle x^2=e_X \, \forall x \in X \Rightarrow x = x^{-1} \, \forall x \in X$

$\displaystyle \forall x,y \in X:$

$\displaystyle xy=(xy)^{-1}=y^{-1}x^{-1}=yx$

So $\displaystyle X$ is abelian. Is that the same as necessarily abelian?

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Not sure how to go about the second part...
• Oct 25th 2010, 02:53 PM
Dinkydoe
What you proved is that a group X with the given property must be Abelian. This means, there can not exist a group with the same property
that is not Abelian. (that's what they mean with necessarily Abelian)

suppose there's a prime $\displaystyle p\neq 2$ such that $\displaystyle p$ divides $\displaystyle |X|$ then there's an element of order p in X....so?

If it X had less then n generators....then $\displaystyle |X|=$...?
• Oct 25th 2010, 03:04 PM
MichaelMath
For the first part, could I just have said :

$\displaystyle \forall x,y \in X$

$\displaystyle x^2=1$ and $\displaystyle y^2=1$ and $\displaystyle (xy)^2=1$

so: $\displaystyle (xy)^2=x^2y^2$

and: $\displaystyle xy=(xy)^{-1}=y^{-1}x^{-1}=yx$
• Oct 25th 2010, 03:10 PM
Dinkydoe
You could have, but the only important part is:

$\displaystyle (\forall x)(x=x^{-1})$

This observation is made by multiplying $\displaystyle x^2=1$ with $\displaystyle x^{-1}$ on both sides...

Hence $\displaystyle (\forall x)(\forall y)(xy = (xy)^{-1}= y^{-1}x^{-1}=yx)$
• Oct 25th 2010, 04:12 PM
MichaelMath
Quote:

Originally Posted by Dinkydoe
suppose there's a prime $\displaystyle p\neq 2$ such that $\displaystyle p$ divides $\displaystyle |X|$ then there's an element of order p in X....so?

If it X had less then n generators....then $\displaystyle |X|=$...?

arghhh... I know it's easy but I'm still not getting this stuff.
• Oct 25th 2010, 04:25 PM
Dinkydoe
If $\displaystyle p\neq 2$ divides $\displaystyle |X|$ then there's an element of order $\displaystyle p$ in $\displaystyle X$....wich is a contradiction (why?)

So $\displaystyle p$ can not divide $\displaystyle |X|$ and consequently, only 2 divides $\displaystyle |X|$ so $\displaystyle |X|=2^n$

If $\displaystyle X$ has $\displaystyle n-k$ generators then $\displaystyle |X|=2^{n-k}...$ (the generators must also be of order 2...)
• Oct 26th 2010, 09:49 AM
Bruno J.
Quote:

Originally Posted by Dinkydoe
If $\displaystyle p\neq 2$ divides $\displaystyle |X|$ then there's an element of order $\displaystyle p$ in $\displaystyle X$....wich is a contradiction (why?)

So $\displaystyle p$ can not divide $\displaystyle |X|$ and consequently, only 2 divides $\displaystyle |X|$ so $\displaystyle |X|=2^n$

If $\displaystyle X$ has $\displaystyle n-k$ generators then $\displaystyle |X|=2^{n-k}...$ (the generators must also be of order 2...)

This is perfect, but you're using a fairly advanced theorem (that there exists an element of order $\displaystyle p$ for a prime $\displaystyle p$ dividing $\displaystyle |G|$).

Another way to see it is that $\displaystyle G$ can be considered as a vector space over the field with two elements. Switching to additive notation, the above identities read as $\displaystyle 2(x+y)=2x+2y$... since every vector space has a basis, we have an isomorphism $\displaystyle G \cong (Z_2)^n$ for some $\displaystyle n$, hence $\displaystyle |G|=2^n$.