Derived series of the Dihedral group

Asked Jan 11 '19 at 19:01

Active Jan 11 '19 at 19:15

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I'm working on derived subgroups because I'm studying for an exam and I want to show that in the case of the dihedral group $D_{2n}=\langle\sigma ,\tau|\sigma^n=\tau^2,\sigma^{\tau}=\sigma^{-1}\rangle$ that the derived series is

$1 \unlhd\langle\sigma\rangle\unlhd D_{2n}$ if n is odd and $1 \unlhd\langle\sigma^2\rangle\unlhd D_{2n}$ if n is even.

I know that the commutator will give $[\sigma,\tau]=\sigma^{-1}\tau^{-1}\sigma\tau=\sigma^{-2}$ which implies that the commutator can always be generated by just sigma , but I am confused as to show the difference between the even and odd cases . any suggestions anyone ?

edited Jan 11 '19 at 19:15

user3482749

6,750

asked Jan 11 '19 at 19:01

excalibirr

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The commutator subgroup is given by $\langle \sigma^2\rangle$. – Dietrich Burde Jan 11 '19 at 19:15
$\sigma$ does not generate the commutator subgroup: it's not even contained in it. – user3482749 Jan 11 '19 at 19:15
@user3482749 In a past exam paper at my college we were asked to show the statement I asked about though ?? – excalibirr Jan 11 '19 at 19:34
No problem, you have your answer already. It is $\langle \sigma^2\rangle$. The first series is not a derived series. – Dietrich Burde Jan 11 '19 at 19:47
@DietrichBurde so do you think it was a trick question , designed perhaps to see if you realise that one of the series is impossible ? – excalibirr Jan 11 '19 at 19:53
If n is odd, $\sigma^2$ has an order n hence the derived group is $Z_n$ which can also be generated by $\sigma$ – Grace Jun 06 '23 at 06:40

Derived series of the Dihedral group

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