Help understanding two Group questions

Problem 1: Suppose *G* is a group with the property that *agb=cgd* always implies that *ab=cd*. Prove that *G* is abelian.

What does this question mean? Is *g* a special element of *G,* or any old element? Suppose *g* is the identity (in which case *ab=cd* anyway). How does one prove this?

Problem 2: Suppose *G* is a finite abelian group in which no non-identity is equal to its own inverse. Determine the product of all the elements of *G*.

How can I do the entire table? It could have thousands (literally!) of entries (or more). How do I know what *a* time *b* equals? I don't get it.

Re: Help understanding two Group questions

1) it's "any old element". the trick is to pick the right choices for a,b,c and d, so that we wind up with two products agb and agd we KNOW are equal. there are different ways to do this, here is one:

choose a = x, b = y, c = y, d = x, and g = x^{-1}

then agb = x(x^{-1})y = (xx^{-1})y = y, and:

cgd = y(x^{-1})x = y(xx^{-1}) = y, so we have agb = cgd. because in G agd = cgd implies ab = cd, we have: xy = yx, so G is abelian.

2) hint: if no element except the identity is its own inverse, multiply all the elements of G together like so:

(e)(g_{1}g_{1}^{-1})(g_{2}g_{2}^{-1})....... = ?

(of course this means that |G| must be odd. why is this not a problem?)

Re: Help understanding two Group questions

Quote:

Originally Posted by

**Deveno** ...choose a = x, b = y, c = y, d = x, and g = x^{-1}

then agb = x(x^{-1})y = (xx^{-1})y = y, and:

cgd = y(x^{-1})x = y(xx^{-1}) = y, so we have agb = cgd. because in G agd = cgd implies ab = cd, we have: xy = yx, so G is abelian.

Good clear answer. But *how is a student supposed to come up with that?* I mean, now that I see it, it's easy to understand. But how does one recognize this?

Is there anyone out there who, as a student, figured out this answer without asking for help, or at least very little help? What was the process by which you got there?

I now need to prove or disprove if the converse is true, so I'm going to try that on my own.

Quote:

Originally Posted by

**Deveno** ...(e)(g_{1}g_{1}^{-1})(g_{2}g_{2}^{-1})....... = ?

OK. I misunderstood the question to mean write the multiplication table of all the elements (duh!). Answer is the identity (which still took me some time to recognize <smh at myself>).

Quote:

Originally Posted by

**Deveno** (of course this means that |G| must be odd. why is this not a problem?)

From the property that non-identities cannot equal their inverse (and your writing it out) it's pretty clear that the elements of G occur in pairs, g and g^{-1}. Thus, there are an even number of non-identity elements. If there is only one identity (and identities are unique, I'm nearly certain), making an odd number of elements to G.

That's not a problem because I have no issues with a finite group having an odd number of elements. Clearly such groups exist, e.g. $\displaystyle (\mathbb{Z}_5,+)$.

Re: Help understanding two Group questions

well, Z4 has an even number of elements (0,1,2, and 3). the point is, the condition that x = x^{-1} is equivalent to saying x is of order 1 or 2. if x ≠ e, then x is of order 2. it turns out that |G| is even iff G has an element of order 2 (one direction is lagrange's theorem, the other is cauchy's theorem for p = 2). if G has elements of order 2, then this (elementary) argument fails.

to answer your question: how would somebody get the first question, if they didn't know the answer?

often, it's by working backwards. you'd like to prove that:

xy = yx

from xgy = ygx, so you need to pick g so that xgy = ygx will always be true. well if xg...(something) = yg...(something else) some sort of cancellation has to be going on, that is: g ought to start with either x^{-1} or y^{-1}.

Re: Help understanding two Group questions

Quote:

Originally Posted by

**Deveno** well, Z4 has an even number of elements (0,1,2, and 3)...

I meant Z5 (and edited my post to make me look less dumb). :D

Thanks for your help.

Re: Help understanding two Group questions

Quote:

Originally Posted by

**Deveno** ...often, it's by working backwards. you'd like to prove that:

xy = yx

from xgy = ygx, so you need to pick g so that xgy = ygx will always be true. well if xg...(something) = yg...(something else) some sort of cancellation has to be going on, that is: g ought to start with either x^{-1} or y^{-1}.

What I missed is fitting the *xy=yx* into *ab=cd*. Possibly if I'd realized that *a=d* and *b=c* I'd get the answer, I could have worked backwards. I need to work on fitting the needed conclusion (*xy=yx*) into some part of the given info.