# Commutators, sort of.

• Oct 26th 2007, 10:39 PM
Soltras
Commutators, sort of.
All right here's my latest one that's already consumed a healthy stack of papers trying to work out. This one is on p.106 of Hungerford's Algebra text, #3(a).

If H and K are subgroups of G let (H,K) be the subgroup of G generated by the elements { $\displaystyle hkh^{-1}k^{-1} : h \in H, k \in K$ }. Show that (H,K) is normal in H v K.

(H v K is the join of H and K, the subgroup generated by elements in the union of H and K).

Here's the progress I've made. First of all, obviously $\displaystyle (H,K) \subset H \vee K$. In fact, it's a subset of the commutator subgroup of $\displaystyle H \vee K$. If I can show containment the other way (which I'm not even sure is correct) then normal property follows because commutator subgroups are normal.
Instead, I've tried letting $\displaystyle x \in H \vee K$ and looking at $\displaystyle x(H,K)x^{-1}$. It should be enough to show that $\displaystyle (xhx^{-1})(xkx^{-1})(xh^{-1}x^{-1})(xk^{-1}x^{-1}) \in (H,K)$, which would be easy if N and K are normal in $\displaystyle N \vee K$, but this isn't given, so it's been difficult. Surely I'm just overlooking some proposition, which is why I'm seeking help here from others who may be more comfortable with group theory. Any help would be appreciated.
• Oct 27th 2007, 04:35 PM
ThePerfectHacker
Quote:

Originally Posted by Soltras
All right here's my latest one that's already consumed a healthy stack of papers trying to work out. This one is on p.106 of Hungerford's Algebra text, #3(a).

If H and K are subgroups of G let (H,K) be the subgroup of G generated by the elements { $\displaystyle hkh^{-1}k^{-1} : h \in H, k \in K$ }. Show that (H,K) is normal in H v K.

(H v K is the join of H and K, the subgroup generated by elements in the union of H and K).

Here's the progress I've made. First of all, obviously $\displaystyle (H,K) \subset H \vee K$. In fact, it's a subset of the commutator subgroup of $\displaystyle H \vee K$. If I can show containment the other way (which I'm not even sure is correct) then normal property follows because commutator subgroups are normal.
Instead, I've tried letting $\displaystyle x \in H \vee K$ and looking at $\displaystyle x(H,K)x^{-1}$. It should be enough to show that $\displaystyle (xhx^{-1})(xkx^{-1})(xh^{-1}x^{-1})(xk^{-1}x^{-1}) \in (H,K)$, which would be easy if N and K are normal in $\displaystyle N \vee K$, but this isn't given, so it's been difficult. Surely I'm just overlooking some proposition, which is why I'm seeking help here from others who may be more comfortable with group theory. Any help would be appreciated.

So let $\displaystyle C = \left< \{ hkh^{-1}k^{-1}\} \right>$ be the generated group of commutators from $\displaystyle H$ and $\displaystyle K$. Let $\displaystyle H \ \bold{v} \ K$ be the smallest group containing $\displaystyle H$ and $\displaystyle K$ (which is exactly what you said but stated differently). It should be trivial that $\displaystyle C\subseteq H \ \bold{v} \ K$. So it is a subgroup it remains to show that it is normal. We need to show $\displaystyle pxp^{-1} \in C$ for all $\displaystyle x\in C$ and $\displaystyle p$ in the join of $\displaystyle H,K$. But before we prove this we will prove a slightly weaker result. That if $\displaystyle x\in C$ is a product of a single commutator then $\displaystyle pxp^{-1} \in C$. So we need to show $\displaystyle p(hkh^{-1}k^{-1})p^{-1}\in C$. Note, $\displaystyle p(hkh^{-1}k^{-1})p^{-1} =( phkh^{-1})(1)(k^{-1}p^{-1})=(phkh^{-1})(p^{-1}k^{-1}kp)(k^{-1}p^{-1})$ rewrite as $\displaystyle [(ph)(k)(ph)^{-1}k^{-1}][kpk^{-1}p^{-1}]$. Now if $\displaystyle p \in H$ (it is either in one of them) then the first factor in that product is a commutatator because $\displaystyle ph\in H$ and so it the second factor (because its inverse is $\displaystyle pkp^{-1}k$ which is a commutator so its inverse is in $\displaystyle C$). But if $\displaystyle p$ is not in $\displaystyle H$ then it must be in $\displaystyle K$, but there is nothing to worry about because $\displaystyle C$ contains products $\displaystyle khk^{-1}h^{-1}$ too, so then this is also in $\displaystyle C$. Hence we have shown that single commutators are in $\displaystyle C$ when we preform conjugation on them. However, if we have several products $\displaystyle c_1c_2...c_k$ where $\displaystyle c_i \in C$ then $\displaystyle pc_1c_2...c_kp^{-1} = (pc_1p^{-1})(pc_2p^{-1})...(pc_kp^{-1})$ and we proved just above that each $\displaystyle pc_ip^{-1}\in C$. So then is their product. Hence it is normal.

By the way, is this the GTM Herstein or UTM Herstein?

EDIT: I want to explain again why $\displaystyle (ph)(k)(ph)^{-1}k^{-1}\in C$ to be more clear. First, if $\displaystyle p \in H$ then it is finished. If not then $\displaystyle p\in K$. Notice that if $\displaystyle hkh^{-1}k^{-1}\in C$ if and only if its inverse $\displaystyle khk^{-1}h^{-1}\in C$ (i.e. $\displaystyle h,k$ are reversed). So it is happened that $\displaystyle p \not \in H$ then repeating the same argument from the begining but instead of $\displaystyle p(hkh^{-1}k^{-1})p^{-1}$ but with them reversed, i.e. $\displaystyle p(khk^{-1}h^{-1})p^{-1}$ we arrived at (the same approach) $\displaystyle (pk)(h)(pk)^{-1}h^{-1}$ which is a commutator of the form $\displaystyle khk^{-1}h^{-1}$ so it is in $\displaystyle C$.
• Oct 27th 2007, 06:58 PM
Soltras
Quote:

Originally Posted by ThePerfectHacker
By the way, is this the GTM Herstein or UTM Herstein?

It's GTM Hungerford, but I also have Herstein's undergrad text, "Topics in Algebra."

Thank you for the proof! You are very helpful. The crux seems to to inserting the identity at various places inside the term to "split" it up and thereby change its form to one we want.

The only step I do not fully comprehend is the part that assumes if p is not in H, then it must be in K. Isn't it true that H v K may include elements not in either H or K, i.e., some products made by combining elements chosen from H-K and K-H. Or are we repeating the same trick we did above for elements of (H,K), which is that if it holds for the generating elements then it holds for all products thereof?
Actually, it seems that is in fact the case, because p is just some combination of elements from H and K, and p^-1 is the combination in reverse order. So that does make some sense, because you can prove it for the "innermost" generating element (which IS in H or K) and work your way outward until you've proved it for p. Thank you so much Hacker.
• Oct 27th 2007, 07:26 PM
ThePerfectHacker
Quote:

Originally Posted by Soltras
It's GTM Hungerford, but I also have Herstein's undergrad text, "Topics in Algebra."

I meant to say ask not Herstein, I ask because in my college book store I saw a Hungerford lying around but it was an easy version.

Quote:

The only step I do not fully comprehend is the part that assumes if p is not in H, then it must be in K. Isn't it true that H v K may include elements not in either H or K.
Yes! I forgot about that :o. But the prove is easily correctable (I hope). I have shown the following that $\displaystyle p(hkh^{-1}k^{-1})p^{-1}$ lies in $\displaystyle C$ if $\displaystyle p\in H \mbox{ or }p\in K$. (You agree with that?). But that does not complete the proof because it can still be the case that $\displaystyle p\not \in (H\cup K)$ (like you said). But since the join is generated by $\displaystyle (H\cup K)$ it means $\displaystyle p=p_1p_2...p_n$ where $\displaystyle p_k \in (H\cup K)$. So $\displaystyle (p_1...p_n)(hkh^{-1}k^{-1})(p_n^{-1}...p_1^{-1})$. Now work from the middle to outside: $\displaystyle (p_1...p_{n-1})(p_nhkh^{-1}k^{-1}p_n^{-1})(p_{n-1}^{-1}...p_{1}^{-1})$. But the middle term is in the $\displaystyle C$ because $\displaystyle p_n\in (H\cup K)$ so it is a commutator group. If $\displaystyle x$ represented that middle term then going to the next middle term we have $\displaystyle p_{n-1}xp_{n-1}^{-1}$ but we have already proven that in general conjugation leaves elements in the commutator (that was the trick with inserting $\displaystyle pp^{-1}$ in between terms). Thus it is again in the commutator group. Keep on doing it until you arrive at the end. Is that better?

EDIT: Reading your last line in the post you seem to be saying the same thing what I just wrote.
• Oct 27th 2007, 07:34 PM
Soltras
Thanks again Hacker, you're really something.
• Oct 27th 2007, 07:40 PM
ThePerfectHacker
If you ever have free time and you want to review your algebra from the beginning I would recommend Beachy and Blair. I first learned algebra from Fraliegh but reading over this book B&B do a far more fantastic job in writing proofs, plus it is more advanced than average algebra books. It has so many detailed examples in it you never seen so much before. And the excercises are excellent too.
• Oct 27th 2007, 08:02 PM
Soltras
Quote:

Originally Posted by ThePerfectHacker
If you ever have free time and you want to review your algebra from the beginning I would recommend Beachy and Blair. I first learned algebra from Fraliegh but reading over this book B&B do a far more fantastic job in writing proofs, plus it is more advanced than average algebra books. It has so many detailed examples in it you never seen so much before. And the excercises are excellent too.

It's funny you should mention that. William Blair was a former student of my professor, so we used that text for undergraduate algebra -- and Herstein was my professor's PhD advisor in Chicago, which is why I bought that text as a supplement, because of how frequently my professor uses it as a source. Funny you mention both Blair and Herstein in one thread.

I actually found an error in Beachy & Blair once, in their definition of cyclotomic polynomials. By their definition (on p.392 of 3rd edition), $\displaystyle \phi_1(x)$ is an empty product. Not very important, but considering how pedantic they usually are in their definitions, it's an oversight!
• Nov 1st 2007, 06:11 PM
ThePerfectHacker
Is your professor by any chance Lance W. Small . How do I know? I look at his math geneology. Look under Herstein but it would be easier to look above William Blair.

If you are not familar with the Math Geneology Project you should look at their site. It does back all the way to the days of Newton and von Leibniz.
• Nov 1st 2007, 08:02 PM
Soltras
Quote:

Originally Posted by ThePerfectHacker
Is your professor by any chance Lance W. Small . How do I know? I look at his math geneology. Look under Herstein but it would be easier to look above William Blair.

If you are not familar with the Math Geneology Project you should look at their site. It does back all the way to the days of Newton and von Leibniz.

Yes, he is my professor. Good call -- nice site.
I knew E. Artin and Zorn were in his mathematical ancestry, but Holder, Kummer, and Gauss are nice pluses. :)