# fun with "necessarily" abelian

• Oct 25th 2010, 03:35 PM
MichaelMath
fun with "necessarily" abelian
I have this problem:

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

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

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For the first part I did:

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

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

$\forall x,y \in X:$

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

So $X$ 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, 03:53 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 $p\neq 2$ such that $p$ divides $|X|$ then there's an element of order p in X....so?

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

$\forall x,y \in X$

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

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

and: $xy=(xy)^{-1}=y^{-1}x^{-1}=yx$
• Oct 25th 2010, 04:10 PM
Dinkydoe
You could have, but the only important part is:

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

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

Hence $(\forall x)(\forall y)(xy = (xy)^{-1}= y^{-1}x^{-1}=yx)$
• Oct 25th 2010, 05:12 PM
MichaelMath
Quote:

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

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

arghhh... I know it's easy but I'm still not getting this stuff.
• Oct 25th 2010, 05:25 PM
Dinkydoe
If $p\neq 2$ divides $|X|$ then there's an element of order $p$ in $X$....wich is a contradiction (why?)

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

If $X$ has $n-k$ generators then $|X|=2^{n-k}...$ (the generators must also be of order 2...)
• Oct 26th 2010, 10:49 AM
Bruno J.
Quote:

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

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

If $X$ has $n-k$ generators then $|X|=2^{n-k}...$ (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 $p$ for a prime $p$ dividing $|G|$).

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