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Finding Composition Series of Groups

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Poirot

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Feb 15, 2012
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My sum total of knowledge of composition series is: the definition, the jordan holder theorem and the fact that the product of the indices must equal the order of the group.
With this in mind, can someone help with me with finding a composition series for the following:


(1) Z60

(2) D12 (dihedral group)

(3) S10 (symmetric group)

I am not looking for just an answer but actually how to go about finding a series.
 

Opalg

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Feb 7, 2012
2,703
My sum total of knowledge of composition series is: the definition, the jordan holder theorem and the fact that the product of the indices must equal the order of the group.
With this in mind, can someone help with me with finding a composition series for the following:


(1) Z60

(2) D12 (dihedral group)

(3) S10 (symmetric group)

I am not looking for just an answer but actually how to go about finding a series.
As a general strategy, start with the whole group, look for a maximal normal subgroup. Then repeat the process.

(1) should be easy, because $\mathbb{Z}_{60}$ is abelian and so every subgroup is normal. You start by looking for a maximal subgroup. For example, you could take the subgroup generated by 2, which is (isomorphic to) $\mathbb{Z}_{30}.$ Now repeat the process: find a maximal subgroup of $\mathbb{Z}_{30}.$ And so on.

For (2), any dihedral group $D_{2n}$ has a subgroup of index 2 (therefore necessarily normal), consisting of all the rotations in $D_{2n}$ and isomorphic to $\mathbb{Z}_{n}$. That subgroup is abelian, so you can proceed as in (1).

For (3), here's a hint.
 
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Poirot

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Feb 15, 2012
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Ok thanks, it seems it should be quite easy but I have across an example which I don't understand. I am told that {0},<12>,<4>,Z48 is a composition series of Z48 but couldn't <24> be inserted between {0} and <12> since it is of order 2 in Z48? (sorry don't know how to do triangles denoting normal subgroup)
 

Opalg

MHB Oldtimer
Staff member
Feb 7, 2012
2,703
Ok thanks, it seems it should be quite easy but I have across an example which I don't understand. I am told that {0},<12>,<4>,Z48 is a composition series of Z48 but couldn't <24> be inserted between {0} and <12> since it is of order 2 in Z48? (sorry don't know how to do triangles denoting normal subgroup)
What you say is quite correct. The standard definition of a composition series requires that each component should be maximal normal in the next one. The series $\{0\}\lhd \langle12\rangle \lhd \langle4\rangle \lhd \mathbb{Z}_{48}$ fails that test on two counts. You could put $\langle24\rangle$ between $\{0\}$ and $\langle12\rangle$; and you could put $\langle2\rangle$ between $\langle4\rangle$ and $\mathbb{Z}_{48}$.
 
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Poirot

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Feb 15, 2012
250
oh sorry, I misread the text (it just said it was a normal series, not maximal). Well I will go away and follow your hints and advice. By the way, in dihedral groups, is it the case that the rotations are in a conjugacy class on their own?, as this would definitly ease the burden working out if a group is closed under conjugacy. Also, are they any rules about what the subgroups of a dihedral group are. Thanks again, and also I would like to say (doubt this is particularly controversial) that I think you are the best mathematician on this site.
 

Deveno

Well-known member
MHB Math Scholar
Feb 15, 2012
1,967
oh sorry, I misread the text (it just said it was a normal series, not maximal). Well I will go away and follow your hints and advice. By the way, in dihedral groups, is it the case that the rotations are in a conjugacy class on their own?, as this would definitly ease the burden working out if a group is closed under conjugacy. Also, are they any rules about what the subgroups of a dihedral group are. Thanks again, and also I would like to say (doubt this is particularly controversial) that I think you are the best mathematician on this site.
the conjugacy class of a rotation will always contain just other rotations (because the rotations form a normal subgroup of the dihedral group), but not all rotations are conjugate (necessarily).

the reason for this is that for some n, D2n may have a non-trivial center, and central elements only contain themselves in their conjugacy class.

for example, in D8 (the symmetries of a square), we have:

[1] = {1}
[r] = {r,r3}
[r2] = {r2} (where the square brackets mean the conjugacy class of an element).

even if we have a trivial center (like with D10 the symmetries of a regular pentagon), we still have:

(rk)r(rk)-1= r (for k = 0,1,2,3,4)
(rks)r(rks)-1 = (rks)r(rks) = (rk)(sr)(rks) = (rk)(r4s)(rks) = (rk)(r4s)(sr-k) = r4

which shows that [r] = {r,r4}.

that is, conjugacy classes of a normal subgroup may still form a (non-trivial) partition of that normal subgroup (as another example, two cycles of the same cycle type may be conjugate in Sn, but not be conjugate in An​).