# Homework Help: Herstein algebra hint(s) desired

1. Jan 13, 2008

### Mathdope

[SOLVED] Herstein algebra hint(s) desired

1. The problem statement, all variables and given/known data
Let G be an abelian group of order n (even), $a_1$ ,..., $a_n$ its elements. Let x = $a_1$ $\cdot$ ... $\cdot$ $a_n$. Show that if G has more than one element b which is not the identity e, such that $b^2 = e$, then x = e.

(from Herstein, Abstract Algebra, section 2.4 problem 43b)
2. Relevant equations
We know that $x^2 = e$.

3. The attempt at a solution
We can assume that b = $a_1$ , $a_2$ and note that we must also have a third solution b = $a_1 \cdot \ a_2$ as well which differs from the first two. It would seem that a reasonable strategy would be to attempt to find a solution to cx = c, but I'm afraid I don't see it (and not for lack of experimentation).

Last edited: Jan 13, 2008
2. Jan 13, 2008

### Jimmy Snyder

What is $a_1 \cdot a_2 \cdot b$? What is x if the order of the group is odd?

3. Jan 13, 2008

### Mathdope

It's e, clearly.

Also, e, pretty clearly, but the order of the group is even in this case.

Do the two questions answered together give the solution? From the sound of your post they do, but I'd be grateful if you could clarify. Thanks much.

4. Jan 13, 2008

### Mystic998

Well, if ${a_i}^2$ isn't the identity, $a_i$ has to have an inverse in the list, right? So isn't x really the product of all elements b with the property $b^2 = e$?

Last edited: Jan 13, 2008
5. Jan 13, 2008

### Mathdope

Absolutely. Apparently any two that are distinct will combine into a third, and the product of those three will always form e. The question then becomes how to determine that "two or more" turns into "exactly enough" for the product of all such elements to give e.

Edit: by the way, it's not clear to me that Lagrange's theorem applies here, but the problem is in the section following Lagrange's theorem and a few of it's simple applications. Just thought I'd mention.

Last edited: Jan 13, 2008
6. Jan 13, 2008

### morphism

Since I didn't know how to give an appropriate hint, here's an outline of (what I think is) the solution. I split it into 4 bits that you can highlight to read.

1. Suppose x isn't equal to e. Then we can write $x=a_1a_2 \ldots a_m$, where each a_i is a distinct nonidentity element of order 2, and m is the smallest positive integer we can get after we do all the possible cancellations.

2. Note that a_i * a_j (i$\neq$j) must be an element of order 2 as well (prove this), so it's equal to some other a_k. Thus m=1 or m=2.

3. In an abelian group G of even order, the set H={g in G: g^2=e} forms a subgroup of even order (why?); in particular, there must be an odd number of non-identity elements of order 2 in G.

4. So m=1, but this contradicts our premise that G has more than one element of order 2.

Another approach is to use the classification theorem for finite abelian groups, but that could be considered too powerful I guess.

Edit2: OK, fixed things up a bit. I'm implicitly using Cauchy's theorem. Did you cover that yet?

Last edited: Jan 13, 2008
7. Jan 13, 2008

### Mathdope

I am not taking a class. Just personal self teaching. That theorem is later in the text, so in theory it shouldn't be required.
It's 3 sections after this one in the text. I think your outline is sound from what I've read...I just have to think about the m = 1 or 2 part a little bit more, but it seems that it must be correct (since if there are more than 3 factors 3 of them have to multiply to e). How are you implicitly applying Cauchy's Theorem (p prime, o(G)= p k, implies G has an element of order p, correct?)

Thanks for your help, it's very much appreciated.

8. Jan 13, 2008

### morphism

I'm applying http://planetmath.org/encyclopedia/CauchysTheoremForFiniteGroups.html [Broken] to guarantee that H has even order (because if an odd prime divides |H| then H is going to have an element a that isn't going to satisfy a^2=1), but I guess you can avoid this if you work hard enough! Although in Herstein's other book, Topics in Algebra, he does Cauchy's theorem for abelian groups particularly early IIRC.

Last edited by a moderator: May 3, 2017
9. Jan 13, 2008

### Dick

Let Z2 be the cyclic group of order two. Now try the argument on Z2^3. There are seven non-identity factors in x, H is the whole group. None of them cancel pairwise. This makes me wonder what the "all possible cancellations" clause means exactly. I don't see why m=1 or m=2?

10. Jan 13, 2008

### morphism

I guess what I was trying to say is that whenever m>=3, then we can cancel 3 factors off. This is true in Z2^3 too, e.g. if we use the carteisan notation we have (a,1,1)*(1,b,1)=(a,b,1), so we can cancel (a,1,1), (1,b,1) and (a,b,1) off. This won't affect the other elements.

11. Jan 13, 2008

### Dick

Still seems a little shady. If I've got a_1...a_m and I want to cancel a1*a2, how do I know a1*a2 is in the remaining list? Maybe it already cancelled against something else?

12. Jan 13, 2008

### morphism

Good point. It looks like the classification theorem it is!

13. Jan 13, 2008

### Dick

Still seems like there ought to be some simple direct argument. I can't see it. Annoying.

14. Jan 13, 2008

### Mathdope

Well, a1*a2 is different from both a1 and a2. Thus if a1*...*aM is the product of all such elements (without cancellation) then we can rewrite it to be a4...aM (after an inconsequential relabeling of a3 = a1*a2), no? Now, repeat the argument until there are 1 or 2 a's remaining (actually it would seem you could only have one remaining since if there are two remaining then their product would have to be present as well).

Last edited: Jan 13, 2008
15. Jan 13, 2008

### Mathdope

Yes, I believe you can merely note that since there are elements of order 2 in H, and since H is itself a group, 2 divides Order(H).

Last edited by a moderator: May 3, 2017
16. Jan 13, 2008

### Jimmy Snyder

So, let me change the notation a little bit.
$e = a_1$
$a_2 \cdot a_2 = e$
$a_3 \cdot a_3 = e$
$a_4 = a_2 \cdot a_3$
Then you have agreed that $a_2 \cdot a_3 \cdot a_4 = e$
and of course, that means $f = a_1 \cdot a_2 \cdot a_3 \cdot a_4 = e$
How about $g = a_5 \cdot \cdots \cdot a_n$? How many factors are there? Is it an odd or an even number of factors? What is $f \cdot g$?

17. Jan 13, 2008

### Dick

Sorry if I'm being dense. I think we are with you up to a4. There are an even number of factors in g. Then what?

18. Jan 14, 2008

### Mathdope

n-4
Since n is assumed even n-4 is even.
both g and x

Last edited: Jan 14, 2008
19. Jan 14, 2008

### ircdan

Here is a sketch, I'll let you fill in the details I omit.

Suppose G has more than one element b s.t. b^2 = e and b != e. Say a_1, and a_2 have this property, so they both have order 2, then (a_1a_2)^2 = e and a_1a_2 ! = e so a_1a_2 also has order 2. For clarity, say a_1a_2 = a_3.

So we have at least three distinct elements of order 2: a_1, a_2, and a_1a_2, and we have that
x = a_1a_2...a_n, then

xa_1a_2a_1a_2 = a_1...a_n(a_1a_2a_1a_2) = a_4...a_n, that is,

x = a_4...a_n is a product of n-4+1 = n-3 elements, which is an odd number of elements, including the identity, so x = e.

20. Jan 15, 2008

### Mathdope

I assume you are applying some other result in that last bit. I can definitely see it if those elements (a_4...a_n) form a subgroup (since it would have odd order and (a_4...a_n)^2 = e). Is that what you are saying? Or is it something else that I am completely missing?