Proof showing group is abelian?

[SMA_01]

Proof showing group is abelian?

Homework Statement

Show that every group G with identity e such that x*x=e for all x in G is abelian.

The Attempt at a Solution

I know that Ii have to show that it's commutative. I start by taking x,y in G and then xy is in G, so

x*x=e
y*y=e,
so (x*y)*(x*y)=e

I'm not sure where to go from here to show that it's commutative...any help is appreciated. Thank you.

I'm not sure where to go from here

 

[jbunniii]

So far so good. Now you want to perform some operations to both sides of the equation in order to end up with just x*y on the left hand side.

 

[SMA_01]

@jbunniii- That's the part I'm a bit confused about, would I use the inverses in this case?

 

[jbunniii]

Yes. You will also use a key fact about the inverses of elements in this group. What does x*x = e imply?

 

[SMA_01]

I see what you mean, so x=e? Thank you!

 

[jbunniii]

I see what you mean, so x=e? Thank you!

No...

x*x = e does not necessarily imply that x = e. If it did, then there would only be one element in this group, namely e.

Think about inverses. If x * x = e, then what is the inverse of x?

 

[SMA_01]

1/x?

 

[jbunniii]

What's the definition of the inverse of an element?

 

[SMA_01]

Suppose that x' denotes the inverse of x,

then by definition of inverse x*x'=e. Are you implying that the binary operation is on x with its inverse? Or maybe x is equal to its own inverse? Maybe I'm off on some tangent here sorry...

 

[genericusrnme]

Use the fact that your group operation must be assosiative, then try and use the fact that x*x=e and y*y=e

Do you know what it means for a group to be abelian?

 

[sunjin09]

Homework Statement

Show that every group G with identity e such that x*x=e for all x in G is abelian.

The Attempt at a Solution

I know that Ii have to show that it's commutative. I start by taking x,y in G and then xy is in G, so

x*x=e
y*y=e,
so (x*y)*(x*y)=e

I'm not sure where to go from here to show that it's commutative...any help is appreciated. Thank you.

I'm not sure where to go from here

Basically x=x^-1 for all x in G, so xy=x^-1y^-1=(yx)^-1=yx

 

[jbunniii]

Suppose that x' denotes the inverse of x,

then by definition of inverse x*x'=e. Are you implying that the binary operation is on x with its inverse? Or maybe x is equal to its own inverse? Maybe I'm off on some tangent here sorry...

Right, x*x = e means that x is its own inverse. This is true for every element in the group. So if

(x * y) * (x * y) = e

then multiplying each side on the left by x and applying associativity gives

(x * x) * y * x * y = x * e

and since x * x = e, the term in parentheses vanishes and you're left with

y * x * y = x

Now proceed to the next logical step.