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Let $G$ be a group with exactly $2$ elements of order $6$. Prove that $G$ has a normal subgroup of order $3$.

Since there is an element of order $6$, by Lagrange's Theorem, the order of $G$ must be a multiple of $6$. That means that both $2$ and $3$ are also divisors of the order of $G$, so, again by Cauchy's Theorem, $G$ must contain elements of order $2$ and order $3$ as well, respectively.

I suppose I'm not sure where to proceed from here. How can we use the fact that $G$ has exactly $2$ elements of order $6$ ? Would Sylow Theorems be helpful here at all ? I don't see how - since we don't know the exact order of $G$ here, which is when I'm used to using the Sylow Theorems.

Any help would be appreciated.

Thanks!

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    $\begingroup$ Is $G$ a finite group? $\endgroup$ – sera Dec 31 '19 at 5:41
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    $\begingroup$ The conclusion that $|G|$ is a multiple of six does not follow from Cauchy’s Theorem (which says that if a prime $p$ divides $|G|$, then $G$ has an element of order $p$). It follows from Lagrange’s Theorem. $\endgroup$ – Arturo Magidin Dec 31 '19 at 6:29
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    $\begingroup$ But $G$ is not necessarily a finite group so the discussions about $|G|$ are irrelevant. As others have pointed out, the two elements of order $6$ mus be self-inverse. $\endgroup$ – Derek Holt Dec 31 '19 at 8:57
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First, a characteristic subgroup of a normal subgroup is normal.

Now let $a$ have order $6$. Then $a^5$ also has order $6$. Hence conjugation by any element of $G$ takes $a$ to $a$ or $a^5$, by the assumption of only two elements of order six. Hence $\langle a\rangle $ is invariant under conjugation, hence normal.

Next, $a^2$ has order $3$.

But $\langle a^2\rangle $ is characteristic in $\langle a\rangle $ (since $\langle a\rangle $ is cyclic).

The result follows.

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  • $\begingroup$ Why does $|a^5| = 6$ imply $\langle{a}\rangle$ being the only subgroup of order $6$? $\endgroup$ – Clement Yung Dec 31 '19 at 8:05
  • $\begingroup$ @ClementYung see my edit. $\endgroup$ – Chris Custer Dec 31 '19 at 8:57
  • $\begingroup$ Okay that's very clear. Thank you. $\endgroup$ – Clement Yung Dec 31 '19 at 9:16
  • $\begingroup$ @ClementYung ok. And thanks for pointing that out. $\endgroup$ – Chris Custer Dec 31 '19 at 13:16

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