By Lagrange's, the subgroups have orders $20$ ($Q_{20}$ itself), $10$, $5$, $4$, $2$, and $1$ (the identity subgroup). For example, a subgroup of order $10$ has elements of order $5$ and $2$ by Cauchy's. I have determined the orders of the elements, I'm just not very sure how to find those subgroups. $Q_{20}=\langle a,b:a^{10}=1,b^2=a^5,ba=a^{-1}b\rangle$ Thanks

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    $\begingroup$ Could you link to a definition of the quaternion group of order $20$? I am only familiar with the ones of order a power of $2$. $\endgroup$ – Tobias Kildetoft Aug 21 '13 at 10:44
  • $\begingroup$ $Q_{20}=<a,b:a^{10}=1,b^2=a^5,ba=a^{-1}b>$ $\endgroup$ – Bran Aug 21 '13 at 10:48
  • $\begingroup$ @TobiasKildetoft: GAP tells that, the group is isomorphic with $C_5:C_4$ $\endgroup$ – Mikasa Aug 21 '13 at 10:54
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    $\begingroup$ @DonAntonio That is the way GAP denotes a semidirect product. $\endgroup$ – Tobias Kildetoft Aug 21 '13 at 11:50
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    $\begingroup$ @subzero155: Oh, I see. Great job! and +1 :-) $\endgroup$ – Mikasa Aug 23 '13 at 9:26

An idea:

We take your presentation for the group (but check the added part for $\,b^4\;$. Just to have a standard presentation):

$$Q_{20}=\left\langle a\,,\,b\;;\;a^{10}=b^4=1\;,\;b^2=a^5\;,\;bab^{-1}=a^{-1}\right\rangle\;,\;\;\text{then}$$

$$\begin{align*}H_2:=&\langle a^5\rangle\;\;&\text{ has order}\;\;2\\ H_4:=&\langle b\rangle \;\;&\text{has order}\;\;4\\ H_5:=&\langle a^2\rangle\;\;&\text{has order}\;\;5\\ H_{10}:=&\langle a\rangle\;\;&\text{has order }\;\;10\end{align*}$$

  • $\begingroup$ $b^4=1$ is a direct consequence of $a^{10}=1$ and $b^2=a^5$. These are not all subgroups. $\endgroup$ – Bran Aug 21 '13 at 11:39
  • $\begingroup$ No...I never mentioned nor hinted these are "all" the groups. These are just some examples of subgroups with the orders you mentioned. For example, you can check that also $\,\left|\,\langle ab\rangle\,\right|=4\;$ ... $\endgroup$ – DonAntonio Aug 21 '13 at 11:43
  • $\begingroup$ Yes I understand. Just wandering how I can find them all and make sure that they are all. $\endgroup$ – Bran Aug 21 '13 at 11:47
  • $\begingroup$ I see...I don't know whether there's a standard method to write down the complete subgroup lattice of a general group, even of a finite one. Perhaps in this particular case something can be done. Anyway, check here:fredonia.edu/department/math/facultypages/straight/groups/pdf/… $\endgroup$ – DonAntonio Aug 21 '13 at 11:54
  • $\begingroup$ I don't think there is a standard way of finding them. I guess only by exhaustion. $\endgroup$ – Bran Aug 21 '13 at 12:09

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