Herstein algebra hint(s) desired

[Mathdope]

[SOLVED] Herstein algebra hint(s) desired

Homework Statement

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)

Homework Equations

We know that $x^2 = e$.

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).

 

[Jimmy Snyder]

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.

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

 

[Mathdope]

What is $a_1 \cdot a_2 \cdot b$?

It's e, clearly.

What is x if the order of the group is odd?

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.

 

[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 $b^2 = e$?

 

[Mathdope]

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 $b^2 = e$?

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.

 

[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?

 

[Mathdope]

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

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.

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

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.

 

[morphism]

I'm applying http://planetmath.org/encyclopedia/CauchysTheoremForFiniteGroups.html 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.

 

[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?

 

[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.

 

[Dick]

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.

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 canceled against something else?

 

[morphism]

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

 

[Dick]

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

 

[Mathdope]

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 canceled against something else?

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).

 

[Mathdope]

I'm applying http://planetmath.org/encyclopedia/CauchysTheoremForFiniteGroups.html 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!

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).

 

[Jimmy Snyder]

It's e, clearly.

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 $f \cdot g$?

 

[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?

 

[Mathdope]

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?

n-4

Is it an odd or an even number of factors?

Since n is assumed even n-4 is even.

What is $f \cdot g$?

both g and x

 

[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.

 

[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?

Sorry for seeming stupid about this one.

 

[Dick]

The other ones can't be assumed to form a subgroup. Look at the example of Z2^3. The elements are (0,0,0),(0,0,1),(0,1,0),(1,0,0),(1,1,0),(1,0,1),(0,1,1) and (1,1,1). The group operation is element-by-element addition mod 2. So for example (0,0,1)+(0,1,0)=(0,1,1). The sum of all three is the identity e=(0,0,0). The remaining elements do not form a subgroup. It really helps to have a nontrivial concrete example when you are trying to evaluate some of these 'proofs'.

 

[Dick]

Ok. Try this for a proof outline. Let H be the subgroup of G consisting of all elements of order two plus e. Prove that the order of H is a power of two. Now pick an element a (not e). Look at the decomposition of H into cosets of the subgroup {e,a}. Each coset looks like {x,xa} for some x. And there are a even number of them if the order of the group is greater than two. The product of the two elements in each coset is a. What's the product of all of the elements?

 

[Mathdope]

First, thanks for the reply.

Ok. Try this for a proof outline. Let H be the subgroup of G consisting of all elements of order two plus e. Prove that the order of H is a power of two.

I can see that this would follow from Cauchy's theorem, but that is not *officially* available in this section. If it can be done without it, great. You don't need to tell me how yet, but I'm just wondering if it's possible.

Now pick an element a (not e). Look at the decomposition of H into cosets of the subgroup {e,a}. Each coset looks like {x,xa} for some x. And there are a even number of them if the order of the group is greater than two. The product of the two elements in each coset is a. What's the product of all of the elements?

If H = {e, a_2, ..., a_k}, then the cosets should number k/2 yes? Since they are pairwise disjoint the product of all the elements from each coset is the product of everything in H? And since this would be equivalent to a^(k/2) where k/2 is an even number (since k is a power of 2 with exponent 2 or more) that would do it.

Did I understand you correctly?

 

[Dick]

First, thanks for the reply.


I can see that this would follow from Cauchy's theorem, but that is not *officially* available in this section. If it can be done without it, great. You don't need to tell me how yet, but I'm just wondering if it's possible.


If H = {e, a_2, ..., a_k}, then the cosets should number k/2 yes? Since they are pairwise disjoint the product of all the elements from each coset is the product of everything in H? And since this would be equivalent to a^(k/2) where k/2 is an even number (since k is a power of 2 with exponent 2 or more) that would do it.

Did I understand you correctly?

Yep. That's it exactly. And yes, I want Cauchy's theorem to claim that k/2 is even. Guess I can't think offhand how to prove that without invoking Cauchy.

 

[Mathdope]

Yep. That's it exactly. And yes, I want Cauchy's theorem to claim that k/2 is even. Guess I can't think offhand how to prove that without invoking Cauchy.

Well, if it is possible I guess I don't see how. But it's not unheard of in Baby Herstein to put in problems that are difficult (i.e. practically impossible) and then reintroduce them later after more machinery is in place. Considering that this particular problem was the last problem in the "Harder problems" section...well

BTW, how do we put "SOLVED" in the header? Is that something a moderator does, am I responsible? Being a novice at these forums I guess I'm curious...

(Never mind...I found the tool.)

Thanks to everyone who participated in this thread!

 

[morphism]

Here's how one can prove that k/2 is even (I gathered that k=|H|) without Cauchy. Since G contains more than one element of order 2, say a_2 and a_3, then the following is a subgroup of H: K={e, a_2, a_3, a_2a_3}. So 4 divides |H|, by Lagrange.

Actually, come to think of it, we can partition G using the cosets of K instead of H. Everything then cancels out nicely. Say we have the representatives g_1, g_2, ..., g_[G:K], then
$$x=(g_1 e g_1 a_2 g_1 a_3 g_1 a_2 a_3) \ldots (g_{[G:K]} e g_{[G:K]} a_2 g_{[G:K]} a_3 g_{[G:K]} a_2 a_3)$$
$$=(g_1 g_2 \ldots g_{[G:K]})^4 (a_2 a_3)^{2[G:K]}$$
$$=(g_1 g_2 \ldots g_{[G:K]})^4$$
$$=e,$$
where this last equality comes from canceling out inverses and using $g_i = (g_i)^{-1} k \implies (g_i)^4 = e$ to kill off the remaining g's.

 

[Mathdope]

Got cosets?

Thanks morphism, that is surely within the limits of what Herstein presents up to the point that he gives the exercise.

 

[Dick]

Here's how one can prove that k/2 is even (I gathered that k=|H|) without Cauchy. Since G contains more than one element of order 2, say a_2 and a_3, then the following is a subgroup of H: K={e, a_2, a_3, a_2a_3}. So 4 divides |H|, by Lagrange.

Actually, come to think of it, we can partition G using the cosets of K instead of H. Everything then cancels out nicely. Say we have the representatives g_1, g_2, ..., g_[G:K], then
$$x=(g_1 e g_1 a_2 g_1 a_3 g_1 a_2 a_3) \ldots (g_{[G:K]} e g_{[G:K]} a_2 g_{[G:K]} a_3 g_{[G:K]} a_2 a_3)$$
$$=(g_1 g_2 \ldots g_{[G:K]})^4 (a_2 a_3)^{2[G:K]}$$
$$=(g_1 g_2 \ldots g_{[G:K]})^4$$
$$=e,$$
where this last equality comes from canceling out inverses and using $g_i = (g_i)^{-1} k \implies (g_i)^4 = e$ to kill off the remaining g's.

Duh. Nice one, morphism. I had a subgroup of order 4 staring me in the face and I ignored it.