# proof in modern algebra

• Oct 28th 2012, 02:31 PM
amyw
proof in modern algebra
Prove that (G, *) is an abelian group if an only if f:G--->G defined by f(x)=x^-1 is a homomorphism.

I understand that I need to prove two things.
1) If (G, *) is abelian, then the function is a homomorphism.
2) If the function is a homomorphism, then (G, *) is abelian.

I know that abelian means that for every a, b in G, a*b=b*a.

I also know that a function is a homomorphism if for every a, b in G, f(ab)=f(a)f(b).

I just don't know how to start...
• Oct 28th 2012, 02:47 PM
GJA
Re: proof in modern algebra
Hi amyw,

Looking at the definitions is a good place to start. Perhaps I can get you going on the forward direction: If G is abelian, then f is a homomorphism. Perhaps if you get this part done you can take another shot at the backwards direction.

As you said, we want to prove that if G is abelian, then $\displaystyle f(a*b)=f(a)*f(b)$ for all $\displaystyle a, b\in G,$ where we are using $\displaystyle *$ to denote the group operation. By definition of f we have $\displaystyle f(a*b)=(a*b)^{-1}=b^{-1}*a^{-1}=\ldots = f(a)*f(b).$ I have left a blank for you to fill in. Notice that all we have used was the definition of f. This means to fill in the blank we need to use the assumption that G is abelian.

Let me know if you are able to fill in this blank. After you do give the reverse direction another shot on your own. If you're still stuck I can write again. Good luck!
• Oct 28th 2012, 05:10 PM
amyw
Re: proof in modern algebra
I think it should be:

LHS: f(a*b)=(a*b)-1=b-1*a-1=a-1*b-1 because (G, *) is abelian.

RHS: f(a) *f(b)=a-1*b-1

I don't understand why f(a*b)=b-1*a-1 instead of a-1*b-1...on the left hand side.

Also, how do you get the math writing like it should be? Do you write it in another program and then cut and paste? or is their an equation writer embedded here that I haven't found yet?
• Oct 28th 2012, 06:36 PM
GJA
Re: proof in modern algebra
1) You're correct, $\displaystyle b^{-1}*a^{-1}= a^{-1}* b^{-1}= f(a)* f(b)$ because $\displaystyle G$ is abelian. Nice job!

2) Your follow up question is basically why does $\displaystyle (a*b)^{-1}=b^{-1}* a^{-1}$ and not $\displaystyle a^{-1}*b^{-1}.$ Remember that the inverse of an element $\displaystyle x$ in a group $\displaystyle G$ (whether the group is abelian or not) is, by definition, the element $\displaystyle y\in G$ that satisfies $\displaystyle xy=e=yx$. When this is the case we write $\displaystyle y=x^{-1}.$ To check that the inverse of $\displaystyle ab$ is really $\displaystyle b^{-1}*a^{-1}$ we compute

$\displaystyle (a*b)*(b^{-1}*a^{-1})=a*(b*b^{-1})*a^{-1}=a*e*a^{-1}=a*a^{-1}=e$

and

$\displaystyle (b^{-1}*a^{-1})*(a*b)=b^{-1}*(a^{-1}*a)*b= b^{-1} *e*b=b^{-1}*b=e.$

Since multiplying $\displaystyle a*b$ on the left and right by $\displaystyle b^{-1}*a^{-1}$ produces the identity, it follows from the definition of the inverse that $\displaystyle (a*b)^{-1}=b^{-1}*a^{-1}.$

To see why $\displaystyle a^{-1}*b^{-1}$ is NOT the inverse of $\displaystyle a*b$ in general, try computing

$\displaystyle (a*b)*(a^{-1}*b^{-1}).$

If your group is not abelian, then you can't slide $\displaystyle a^{-1}$ past $\displaystyle b$ to cancel out the $\displaystyle a$ in front. So, unless your group is abelian, the inverse of $\displaystyle a*b$ is in general not $\displaystyle a^{-1}*b^{-1}$.

So, going back to f, we have $\displaystyle f(a*b)=(a*b)^{-1}$ by the definition of $\displaystyle f$ and $\displaystyle (a*b)^{-1}=b^{-1}*a^{-1}$ from what was discussed above. So $\displaystyle f(a*b)=(a*b)^{-1}=b^{-1}*a^{-1}.$

Does that clear things up about inverses and f?

3) To get the nice looking math writing you need to know a computer language called LaTeX. Do you know it? If you do you can click on the "Go Advanced" button when you post, then click the captial sigma. This will provide an environment where you can produce LaTeX in an HTML setting.

Let me know if there's anything that's unclear. Good luck!
• Oct 28th 2012, 10:35 PM
Deveno
Re: proof in modern algebra
in general, inversion is "order-reversing".

it turns out that elements of a group can be thought as standing for "actions". for example, in the integers, "1" can be thought of as both the distance 1, or an instruction to "move one unit" (usually to the left, but this is kinda arbitrary). this kind of instruction is known as "translation", in geometric terms.

so every group operation is, in some sense, "composition of actions". that is: there is an interpretation of any group G in which a*b means: do b, then do a (some texts use the opposite convention, that a*b means do a, then b. which convention we pick is, again, sort of arbitrary. it helps to be consistent once having made this important choice).

well think of putting on your socks, and then putting on your shoes. to "undo" this action, one reverses the order (taking off your socks first would be quite the feat!), and takes off the shoes, THEN the socks.

with some things, it doesn't matter which one we do first, the end result is the same. this is in a sense what "commuting" (abelian-ness, or commutativity) means. for example, if my actions are: "putting things in a box", then they commute, i can take them out of the box in a different order than i put them in, and still "undo" my actions either way. "lumping things together" (adding) is one of the most important kinds of abelian actions, it doesn't matter when i add 2 and 3, which one i start with, and which one i add next, i get 5 either way. so + is usually reserved for abelian groups only.

matrix multiplication is perhaps the first "concrete thing" you meet where objects DON'T commute. and it turns out that MOST groups can be thought of as consisting of matrices. for example here is a cyclic group of order 4:

$\displaystyle G = \left\{ \begin{bmatrix}1&0\\0&1 \end{bmatrix}, \begin{bmatrix}0&-1\\1&0 \end{bmatrix}, \begin{bmatrix}-1&0\\0&-1 \end{bmatrix}, \begin{bmatrix}0&1\\-1&0 \end{bmatrix}\right\}$

you might recognize the 2nd matrix as a rotation through 90 degrees clockwise, that is: "a quarter-turn".
• Oct 28th 2012, 11:02 PM
amyw
Re: proof in modern algebra
2. Now that I think that through, that was a bit of a silly question. I do math better in the am...like 7, not 2.
3. No, I don't, but if I can learn this, I'm hopeful I can learn it. Just not tonight...
4. So, when I'm proving the other way: I assume f(x)=x^-1 is a homomorphism, and show a^-1*b^-1=b^-1*a^-1. So that's easy...assume f(a*b)=f(a) *f(b), show a^-1*b^-1=b^-1*a^-1
LHS: f(a*b)=b^-1*a^-1
RHS; f(a)*f(b)=a^-1*b^-1
.
so, b^-1*a^-1=a^-1*b^-1. Therefore (G, *) is abelian. :) Thanks very much.
• Oct 28th 2012, 11:09 PM
amyw
Re: proof in modern algebra
Thanks. That was helpful in that it made what I'm learning about a bit more concrete. This class is my first experience with the idea of groups and binary operations. Can you believe I made it to 44 years old without knowing the term homomorphism? I will use this explanation with some of my classmates. Most are having much more trouble with this than I am.
• Oct 28th 2012, 11:46 PM
Deveno
Re: proof in modern algebra
well, we learn a lot of stuff about numbers rather early in life. in particular, we learn (for integers, rational numbers (fractions), real numbers, and/or complex numbers):

a+b = b+a
ab = ba

rather early on, and this "instrinct" becomes ingrained.

so when it comes time to learn about "general" binary operations, the fact that we cannot always use a*b = b*a to simplify things, is disorienting. we've gotten used to "too much structure", and feel lost when we have LESS (what are the rules, what is "safe" to do?).

in the general "math world", right and left can make a difference. so we have to keep track "which side we multiply on". with that one precaution, we can use most of our high-school algebra fairly intact.

non-abelian groups are definitely "weirder". so whenever we find out a group is abelian, it's time to celebrate.