2
$\begingroup$

I have difficulty in understanding nilpotent groups, Although I have understood the definition of nilpotent group to some extent. Let the commutator of $x$ and $y$ is:

$$[x,y]= x^{-1}y^{-1}xy$$

$$[G, G] = \langle \{[x, y]\ \mid x,y\in G\}\rangle$$

$[G , G]$ is called the commutator subgroup of group $G$ and it is easy to verify that it is a subgroup and even more that it is a normal subgroup.

Now let us see the lower central series of group $G$.

$$G = L^{0}(G) \ge L^{1}(G) \cdots $$

Where $L^{i+1}(G) = [G,L^{i}(G)]$.

If the lower central series for $G$ terminates in $\{1\}$ then we say group is nilpotent.

For these three questions consider the underlying group is finite.

Question 1 : I can prove that $L^{i} (G) \ge L^{i+1} (G)$, Is there any simple way to visualise why the size is decreasing ( for some groups ) in the series, where size mean number of elements in $L^{i}(G).$

Question 2 : Is the length of central series tell us, how far is our group from being an abelian group? For an example if length of central series is one than our group is abelian, so to me it appears that as the length of lower central series increase our group start unfollowing the properties of Abelian group.

Question 3: Abelian group are easy to study because definition for a group to be abelian is easy, Is there any way to study nilpotent group that makes various claims, proofs about them little easy?

$\endgroup$
  • $\begingroup$ It is probably better to define a nilpotent group as being a group that has a central series. You can prove that the lcs is the fastest descending. For finite $p$-groups it is easer to use the upper central series to prove nilpotency, using the fact that they have nontrivial centres. Also your comment about number of elements in $L^i(G)$ does not make much sense for infinite groups. $\endgroup$ – Derek Holt Feb 23 '18 at 8:31
  • $\begingroup$ @ Derek Holt I somehow only want to use the definition I have written above. One more thing that I am only interested in finite groups. $\endgroup$ – old Feb 23 '18 at 8:50
  • $\begingroup$ Since the lower central series is the fastest descending central series, any group that has a central series is nilpotent by your definition. $\endgroup$ – Derek Holt Feb 23 '18 at 9:41
  • $\begingroup$ Answer $2$: in some sense, yes! That's exactly the point of defining nilpotent groups: you want the next easiest case after abelian ones. If $L^n(G)=\{1\}$, it means exactly that the elements of $L^{n-1}(G)$ commute with all elements of $G$. $\endgroup$ – 57Jimmy Feb 23 '18 at 11:06
  • $\begingroup$ @Derek Holt This does not seems true to me as $S_3$ is not nilpotent, but it admits a lower central series $S_3 \ge C_3 \ge \{1\}$, where $C_3$ is cyclic group of order 3. $C_3$ is of index 2 in $S_3$ so it is normal in $S_3$. Am i missing something here? $\endgroup$ – old Feb 23 '18 at 11:55
1
$\begingroup$
  1. It's not clear to me that your question makes sense. The size will (usually) decrease for a while, and then stabilise (i.e. stop decreasing and stay the same forever). If it eventually hits the trivial group, then your original group was nilpotent (that's just the definition of nilpotent). But, as I demonstrated in a comment above, $S_3$ is not nilpotent, because its lower central series is $S_3 \geq C_3 \geq C_3 \geq C_3 \geq \dots$, i.e. it's stabilised at something that's not the trivial group.

    Sometimes it even starts off stable and never decreases at all: for example, the lower central series of $A_5$ is $A_5 \geq A_5 \geq A_5 \geq \dots$.

    On the other hand, a group like the dihedral group $D_8$ is nilpotent: you can check that its lower central series goes $D_8 \geq C_2 \geq \{1\} \geq \{1\} \geq \dots$.
  2. You can think of the length of the lower central series (aka "nilpotency class") of a nilpotent group as one measure of how far the group is from being abelian, yes.

    An abelian group is one in which $xy = yx$ for all $x$ and $y$. On the other hand, a nilpotent group is one in which $xy = yx\varepsilon$ for all $x$ and $y$, where $\varepsilon$ is a kind of 'error term'; this error term might not be $1$ (like in the abelian case), but it should be a step closer to $1$ than both $x$ and $y$ (and your group should be made up of finitely many such 'steps').
  3. Yep - we wouldn't bother to use this definition if it didn't help us! The analogy I drew with abelian groups in the previous paragraph can often be made rigorous and turned into a proof by induction (on how many steps $\varepsilon$ is away from $1$), but of course it depends on what you're trying to prove. See, for example, Derek Robinson's book "A course in the theory of groups", which has a nice section (5.2) on nilpotent groups: some of the proofs here use induction on the nilpotency class.
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.