I am looking to find all subgroups of the following generalised quaternion group $$Q_{20}= \langle a,b: a^{10}=1,b^2=a^5,ba=a^{-1}b\rangle$$ the question asks for all the elements and their orders and conjugacy classes and all subgroups and all normal subgroups.

I got all elements and orders also got all conjugacy classes

I know how to do find all subgroups by writing all possible subsets of orders 2,4,10 but i guess there must be an easy way.

any help appreciated


For the orders $2$ and $5$, you know that groups of these orders must be cyclic, so you build them as $\langle g \rangle$, where the element $g$ has the proper order. You will find that there is one each.

For order $4$, there are the cyclic ones, which you build as above. You will find there are $5$ of these. Another possibility would be a Klein group. But this requires three involutions, and you have shown there is only one.

Finally, as to order $10$, there is the cyclic group $\langle a \rangle$. Since in a non-cyclic group of order $10$ there are $5$ involutions (use Sylow's theorems, for instance), this is the only one.

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    $\begingroup$ A group of prime order is cyclic. An involution is an element of order $2$. $\endgroup$ – Andreas Caranti Feb 23 '16 at 22:24
  • $\begingroup$ for order 4 and 10 how can i prove there is no non cyclic group $\endgroup$ – MRK Feb 23 '16 at 22:26
  • $\begingroup$ You said you have enumerated all elements with their orders. So you should have seen there is only one element of order $2$. $\endgroup$ – Andreas Caranti Feb 23 '16 at 22:29
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    $\begingroup$ OK, if a group of order $4$ has only one element of order $2$, what can the order of the other elements be? $\endgroup$ – Andreas Caranti Feb 23 '16 at 22:33
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    $\begingroup$ If a group of order 10 has only one element $z$ of order $10$, what can the order of the other elements be? If there is an element of order 10, you're done. If, by way of contradiction, there are only elements of order $5$, then the product of $z$ by an element of order $5$ has order $10$. $\endgroup$ – Andreas Caranti Feb 23 '16 at 22:34

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