Abelian Proof

\forall x,y \in G, xy=x^{-1}y^{-1}

, prove that

G

must be abelian.

My proof goes something like this:

xy = x^{-1}y^{-1} \iff x^2y^2 = e

, but note also that

yx = y^{-1}x^{-1} \iff y^2x^2 = e

. So that

x^2y^2 = y^2x^2.

But I have a sinking feeling in my stomach that doesn't prove that

G

is abelian. :frown:

It needs to reduce down to

xy = yx

, doesn't it?

How can I go about this proof? :smile:

Reply 1

Smaug123

Original post by Zacken

\forall x,y \in G, xy=x^{-1}y^{-1}

, prove that

G

must be abelian.

My proof goes something like this:

xy = x^{-1}y^{-1} \iff x^2y^2 = e

, but note also that

yx = y^{-1}x^{-1} \iff y^2x^2 = e

. So that

x^2y^2 = y^2x^2.

But I have a sinking feeling in my stomach that doesn't prove that

G

is abelian. :frown:

It needs to reduce down to

xy = yx

, doesn't it?

How can I go about this proof? :smile:

You're quite right: you've shown that products of squares must commute, but not that products of all elements must commute.

In fact, this kind of group is *very* specific. The first thing you should do when seeing rules like this is to check some special cases. What happens when x=y? When x=y^{-1}? When x=e?

Reply 2

Zacken

Original post by Smaug123

x=y \Rightarrow x^2 = x^{-2}

x=y^{-1} \Rightarrow y^{-1}y = yy^{-1} = e

x=e \Rightarrow y = y^{-1} \Rightarrow y^2 = e

I'm not really seeing how that gives any intuitive help. :redface:

Reply 3

Smaug123

Original post by Zacken

x=y \Rightarrow x^2 = x^{-2}

x=y^{-1} \Rightarrow y^{-1}y = yy^{-1} = e

x=e \Rightarrow y = y^{-1} \Rightarrow y^2 = e

I'm not really seeing how that gives any intuitive help. :redface:

Indeed, the first two do not help. But the third is a really really strong condition on

G

: every element is self-inverse. (That is, G is a "boolean group".) Among other things, if the group is finite, it is therefore of order a power of 2.

I'm trying to think of a non-magic way to show that the group is abelian. The proof follows in one line, but the correct expression to consider is a bit unmotivated. Give me a minute :smile:

Reply 4

Smaug123

Original post by Zacken

x=y \Rightarrow x^2 = x^{-2}

x=y^{-1} \Rightarrow y^{-1}y = yy^{-1} = e

x=e \Rightarrow y = y^{-1} \Rightarrow y^2 = e

I'm not really seeing how that gives any intuitive help. :redface:

OK, motivation: we'll go for a contradiction. Suppose

xy \not = y x

. What can we deduce?

Reply 5

Zacken

Original post by Smaug123

OK, motivation: we'll go for a contradiction. Suppose

xy \not = y x

. What can we deduce?

We can deduce that

y^{-1}xyx^{-1} \neq e

Reply 6

Smaug123

Original post by Zacken

We can deduce that

y^{-1}xyx^{-1} \neq e

And given that

x, y

are self-inverse…?

Reply 7

Zacken

Original post by Smaug123

And given that

x, y

are self-inverse…?

Then

y^{-1}x^{-1}y^{-1}x^{-1} = yxyx = (yx)^2 \neq e

Reply 8

Smaug123

Original post by Zacken

Then

y^{-1}x^{-1}y^{-1}x^{-1} = yxyx = (yx)^2 \neq e

Precisely. That's a contradiction.

More neatly stated (without contradiction):

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

Reply 9

Zacken

Original post by Smaug123

Precisely. That's a contradiction.

More neatly stated (without contradiction):

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

Oh my god! I'm so dense. That was so obvious. Thank you! :eek:

Reply 10

Smaug123

Original post by Zacken

Oh my god! I'm so dense. That was so obvious. Thank you! :eek:

No problem - it's really simple if you spot the right expression to consider, but that takes a bit of intuition. Contradiction often reduces the amount of intuition you need, but tends to produce a less neat proof.

Reply 11

Zacken

Original post by Smaug123

Wait a second, re-reading your proof, I can't see why

xy = (xy)^{-1}

. We're given that

xy = x^{-1}y^{-1}

but

(xy)^{-1} = y^{-1}x^{-1} \neq x^{-1}y^{-1}

unless we assume G is abelian? :confused:

Reply 12

Smaug123

Original post by Zacken

Wait a second, re-reading your proof, I can't see why

xy = (xy)^{-1}

. We're given that

xy = x^{-1}y^{-1}

but

(xy)^{-1} = y^{-1}x^{-1} \neq x^{-1}y^{-1}

unless we assume G is abelian? :confused:

x y \in G

, and we've already shown that for all

g \in G

g = g^{-1}

Reply 13

Zacken

Original post by Smaug123

x y \in G

, and we've already shown that for all

g \in G

g = g^{-1}

Fair enough

(xy)^{-1} \in G

but where does the equality of

xy = (xy)^{-1}

come from?

Reply 14

Smaug123

Original post by Zacken

Fair enough

(xy)^{-1} \in G

but where does the equality of

xy = (xy)^{-1}

come from?

Let

g = xy

. Then

e g = e^{-1} g^{-1}

by the original condition in the question (letting

x = e, y = g

), so

g = g^{-1}

. That is,

(xy) = (xy)^{-1}

Reply 15

Zacken

Original post by Smaug123

Let

g = xy

. Then

e g = e^{-1} g^{-1}

by the original condition in the question (letting

x = e, y = g

), so

g = g^{-1}

. That is,

(xy) = (xy)^{-1}

*wipes sweat off forehead* This abstractness is hard to swallow. Thanks again, it makes a load more sense now. :smile:

+Rep

Reply 16

Smaug123

Original post by Zacken

*wipes sweat off forehead* This abstractness is hard to swallow. Thanks again, it makes a load more sense now. :smile:

+Rep

Thanks

sorry, do tell me if I'm being too abstract. It takes a bit of time to get the right frame of mind to just swap variables in and out for each other like that. A big part of becoming a mathematician is, in my opinion, gaining the ability to detach your understanding of a thing (eg. the element

xy

as a member of

G

) from its name ("

xy

", which is not helpful when it comes to deriving the fact that

xy = (xy)^{-1}

Reply 17

Zacken

Original post by Smaug123

Thanks

xy

as a member of

G

) from its name ("

xy

", which is not helpful when it comes to deriving the fact that

xy = (xy)^{-1}

No, no, you make way more sense than my textbook! :tongue:

- quite a good teacher, really. :smile:

I'm still struggling to get used to that, trying to detach my pre-conceived notions, but I always get scared that detaching them might lead to mistakes somewhere, so I juggle the two in my head. I'll get the hang of it someday. :biggrin: