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I am trying to prove that a group $G$ containing $20$ elements of order $19$ is not cyclic.

With just this information I'm not sure how to go about finding a generator and am looking at proving this more generally. I was thinking of using the fact that a group is not cyclic if there are two subgroups of the same order but not completely sure how to go about that.

Shaun
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slplahn91
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3 Answers3

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Choose an element $x\in G$ of order 19. As the subgroup generated by $x$ has cardinality $|\langle x\rangle |$ = 19 and there are 20 elements of order 19, we find an element $y \in G \setminus \langle x\rangle$ of order 19. Now, $\langle x\rangle$ and $\langle y\rangle$ are two distinct subgroups of $G$ that are both of the same order. Hence $G$ is not cyclic.

Klurp
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For a finite $G$, it follows from McKay’s proof of Cauchy’s Theorem that the number of elements of order $19$ must be $\equiv -1$ mod $19$. Hence the indicated group does not exist.

Nicky Hekster
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  • The question is a bit ambiguous, but it doesn't say "exactly 20 elements", so presumably $G$ is allowed contain more than 20 elements of order 19? –  Feb 14 '20 at 21:04
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    Yes true, I interpreted as exactly $20$ elements of course. – Nicky Hekster Feb 14 '20 at 21:16
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At least $19$ such elements is a sufficient condition to get the claim, as it would imply that there are more than one cyclic subgroup of order $19$ (each containing exactly $18$ elements of order $19$). And elements of order $19$ can come in multiple of $18$, only, because each such an element lie in one and only one subgroup of order $19$.

citadel
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