Order of elements

  • Thread starter Bachelier
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  • #1
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If a group G is non abelian and a and b in G have orders |a|= n and |b|= m, is there a correlation we can draw between m and n and |ab|?
 

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  • #2
Deveno
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in general, no. it may be that |ab| isn't even finite.

for example, if G is the quotient of the free group on two generators defined by:

a2 = b2 = e

then ab is not of finite order, even though both a and b are of order 2.
 
  • #3
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in general, no. it may be that |ab| isn't even finite.

for example, if G is the quotient of the free group on two generators defined by:

a2 = b2 = e

then ab is not of finite order, even though both a and b are of order 2.
Thanks. That's what I thought.

So How do I go about figuring out the order of different elements in a group. For instance, consider Dihedral D9 then I can define as : {a,b| |a|=9 and |b|=2}

Now if I'm trying to figure out the different Sylow 2-subgroups, I need to get the elements with order 2. How do I do it?
 
  • #4
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I guess in this case everything that is multiplied by b^2=e has order 2, it seems like I should treat e as zero under the multiplication rule.
 
  • #5
Deveno
Science Advisor
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Thanks. That's what I thought.

So How do I go about figuring out the order of different elements in a group. For instance, consider Dihedral D9 then I can define as : {a,b| |a|=9 and |b|=2}

Now if I'm trying to figure out the different Sylow 2-subgroups, I need to get the elements with order 2. How do I do it?
for the dihedral group Dn, the elements of order 2 depend on whether or not n is even.

if n is odd, then the sylow 2-subgroups are all of order 2, because half the elements are of odd order (the rotations). geometrically, it's obvious the reflections are of order 2, but we can prove this algebraically, as well:

given: rn = 1, s2 = 1, and rs = sr-1, let's calculate (srk)2.

first, a small detour:

lemma: rs = sr-1 → rks = sr-k.


for k = 1, this is given by the defining relations on Dn.

suppose this is true for k = m.

then rm+1s = r(rms) = r(sr-m) = (rs)r-m
= (sr-1)r-m = sr-(m+1),

so the general result holds by induction on k.

corollary: srk = r-ks

since r-k = rn-k,

we have rn-ks = srk-n, so rnr-ks = srkr-n, so the result follows, because both rn and r-n are the identity.

ok, back to showing what the square of a reflection is:

(srk)2 = (srk)(srk)

= (r-ks)(srk) = r-k(ss)rk

= r-krk = 1.

so every reflection has order 2. now, since 9 is odd, this means we have 9 sylow 2-subgroups, one for each reflection.

(if n is even, we have a slightly more involved situation, because now 4 divides the order of the group).
 

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