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Consider the dihedral group $D_{182}$ of order $182=2\cdot 91$.

I am to show that this group doesn't contain a cyclic subgroup of order $14=2\cdot 7$.

My approach: The textbook states that for a group G of order $pq$ (for primes $p$ and $q$, with $p<q$), if $p$ divides $q-1$, there is a unique non-abelian group of order $pq$.

Since 2 and 7 are primes, and $2<7$, I can show that since 2 divides $7-1$, the subgroup of order 14 must be unique and non-abelian. Since all cyclic groups are abelian, this subgroup can't be cyclic. And since it is unique, no other subgroups of order 14 exist.

Is this correct and sufficient? The exercise hints that information from a completely different chapter can be used, so I wonder if I've missed something.

Thanks in advance!

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  • $\begingroup$ When the text says that there is a unique non-abelian group of that order, it means that any two non-abelian groups of that order are isomorphic to one another. It doesn't say that a group can't contain more than one of these isomorphic groups as subgroups, and it doesn't say that there aren't abelian groups of that order. There are certainly abelian groups of order 14! (IThere are abelian groups of all possible orders.) $\endgroup$
    – John Brevik
    Apr 6, 2015 at 11:12
  • $\begingroup$ It also doesn't say that your group $D_{182}$, which is what you're talking about, actually contains such a subgroup. My guess is that a good way to approach this problem is via the Sylow theorems...but that's only a guess. $\endgroup$
    – John Hughes
    Apr 6, 2015 at 11:16
  • $\begingroup$ Thanks for the clarification John Brevik, I see my mistake! $\endgroup$
    – user229123
    Apr 6, 2015 at 11:20

1 Answer 1

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The dihedral group $D_{2n}$ contains a subgroup $C_n$ of order $n$. All other elements are involutions (i.e., are of order $2$). Therefore any cyclic subgroup of $D_{2n}$ is either of order $2$ or of order dividing $n$.

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  • $\begingroup$ So since 14 doesn't divide 91, it can't be a cyclic subgroup? Thanks! $\endgroup$
    – user229123
    Apr 6, 2015 at 11:30

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