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September 30th, 2017, 06:41 AM  #1 
Member Joined: Jan 2015 From: usa Posts: 67 Thanks: 0  Question in abstract algebra
I need help to answer the following problem: Let $F$ be a field, $n\ge 1$ and $G=F^n\bigoplus F^n\bigoplus Mat(n,F).$ That is an element of $G$ has the form $(v,w,X)$ where $v$ and $w$ are nby1 vectors and $X$ is $n$by$n$ matrix, all with entires in $F$. Define an operator on $G$ by: $$(v_1,w_1,X_1)(v_2,w_2,X_2)=(v_1+v_2,w_1+w_2,X_1+ X_2+v_1w_2^t)$$ Show that $G'=\left\{(0,0,X), X\in Mat(n,F)\right\}$ Thanks in advance. Last edited by greg1313; September 30th, 2017 at 08:18 AM. 
September 30th, 2017, 11:39 AM  #2 
Senior Member Joined: Aug 2012 Posts: 1,572 Thanks: 379 
What is notation G'?

September 30th, 2017, 11:44 AM  #3 
Member Joined: Jan 2015 From: usa Posts: 67 Thanks: 0 
$G'$ is the derived subgroup of $G$ i.e. $Gâ€²=\left\{[a,b],a,bâˆˆG\right\}$
Last edited by skipjack; September 30th, 2017 at 08:29 PM. 
September 30th, 2017, 11:46 AM  #4 
Senior Member Joined: Aug 2012 Posts: 1,572 Thanks: 379 
Ah commutator subgroup. Hadn't heard it called derived subgroup before.
Last edited by Maschke; September 30th, 2017 at 11:50 AM. 
September 30th, 2017, 11:50 AM  #5 
Member Joined: Jan 2015 From: usa Posts: 67 Thanks: 0 
exactly 
September 30th, 2017, 01:29 PM  #6 
Member Joined: Jan 2015 From: usa Posts: 67 Thanks: 0 
You didn't find the answer? I tried, but I didn't find anything.
Last edited by skipjack; September 30th, 2017 at 08:16 PM. 
September 30th, 2017, 02:41 PM  #7 
Senior Member Joined: Aug 2012 Posts: 1,572 Thanks: 379  I'll play around with this later, time and attention span permitting. No guarantees.
Last edited by skipjack; September 30th, 2017 at 08:17 PM. 
September 30th, 2017, 08:27 PM  #8 
Global Moderator Joined: Dec 2006 Posts: 18,053 Thanks: 1395  
September 30th, 2017, 09:48 PM  #9 
Senior Member Joined: Aug 2012 Posts: 1,572 Thanks: 379 
Ok I think I've got half of the required equality at least. For my own sake I'll recap the basics. $G$ is a group and for $x, y \in G$ we define the commutator of $x$ and $y$ as $[x, y] = xyx^{1}y^{1}$. We call an element of $G$ a commutator if it can be expressed as the commutator of some two elements. Clearly if $G$ is Abelian then all commutators are the identity and this is boring. If there aren't too many nontrivial commutators the group is "near" Abelian. If there are a lot of commutators the group is far from being Abelian. If we define the subgroup generated by all the commutators, $G' = \langle [x,y] : x, y \in G\}\rangle$, then the size of $G'$ is a measure of how nonabelian a group is. In fact $G'$ is normal and $G / G'$ is Abelian, called the Abelianization of $G$. Note that the angle brackets indicate not just the commutators, but the collection of all finite products of commutators. It turns out that sometimes these are not the same. The smallest example where they differ is at order 96, where there are two different examples. "Today I learned." Turning to the problem at hand, there's a point of interest in the third coordinate of the product as it's defined. That's the notation $v_1 w_2^t$. Recall that if $v$ and $w$ are $n \times 1$ column vectors, then the matrix product $v^t w$ is a $1 \times 1$ matrix, or scalar, known as the inner product. If you instead compute the matrix product $v w^t$ you get an $n \times n$ matrix called the outer product. So in fact the notation in the problem makes sense, we're just adding an $n \times n$ matrix to another one. In what follows we'll use the fact, verified by easy computation, that $v 0^t = 0v^t$ is the $n \times n$ matrix composed entirely of zeros, or the zero matrix. Now in order to do this problem we have to first figure out what is the identity and what are inverses. * Identity. I claim that $(0, 0, 0)$ is the group identity. In fact, $(v, w, X)(0, 0, 0) = (v, w, X + 0 = X)$ so that's true. * Inverse. The inverse of $(v, w, X)$ is $(v, w, X + vw^t)$. We have $(v, w, X)(v, w, X + vw^t) = (0, 0, X  X + vw^t vw^t = 0)$ Now that we now what inverses look like we can compute the commutator $[(v_1, w_1, X_1), (v_2, w_2, X_2)]$. Before jumping into the computation it's clearer to first write out the inverses. We have: $(v_1, w_1, X_1)^{1} = (v_1, w_1, X_1 + v_1 w_1^t)$ and $(v_2, w_2, X_2)^{1} = (v_2, w_2,  X_2 + v_2 w_2^t)$ Then $[(v_1, w_1, X_1), (v_2, w_2, X_2)]$ $= (v_1, w_1, X_1)(v_2, w_2, X_2) (v_1, w_1, X_1)^{1} (v_2, w_2, X_2)^{1}$ $= (v_1, w_1, X_1)(v_2, w_2, X_2)(v_1, w_1,  X_1 + v_1 w_1^t) (v_2, w_2,  X_2 + v_2 w_2^t)$ The $v$'s and $w$'s all cancel out nicely in the first two coordinates. The third coordinate is some $n \times n$ matrix. We have just proved that the set of commutators is a subset of the set of all elements of the form $(0, 0, X)$. We're not done, because the commutator subgroup is the subgroup generated by all the commutators. That's the set of all finite products of commutators. I convinced myself earlier that any finite product of elements of the form $(0, 0, X)$ has the same form but I'll skip the proof. So we have shown that $G'\subset \{(0,0,X)\}$. That's half of the inequality we are required to prove. Now to go the other way we have to take some arbitrary matrix $X$ and find its decomposition as finite product of commutators. I don't have that worked out yet. I have one concern which is that we never used the fact that $F$ is a field. We only used the properties of a commutative ring. This would all go through over the integers for example. [I'm not even sure I needed commutativity]. It always makes me nervous when the givens are too strong. Either I missed something or the givens really are too strong. Last edited by Maschke; September 30th, 2017 at 10:19 PM. 
September 30th, 2017, 11:08 PM  #10 
Member Joined: Jan 2015 From: usa Posts: 67 Thanks: 0 
thanks you Could you please help me to show that $(0,0,I_n)\in G'$ cannot be expressed as the product of fewer than $[n/2]$ commutators 

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