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I am trying to prove that an abelian group of order 6 has exactly one element of order 2.

I know there is at least one by Cauchy's Theorem, so I am trying to show there are no more than one by contradiction.

Suppose there are $a,b, a \neq b$ such that $a^2 = b^2 = e$. Then also $ab$ has order $2$, so we have two remaining elements (aside from $e, a, b, ab$) of which at least one of them, say $c$, has order $3$ by Cauchy. Then $ac$ has order $6$, but also $bc$ has order $6$ and there is only one element of order 6 (since $e$ has order 1, $a, b, ab$ have order 2, and $c$ has order 3) so $ac = bc$ and therefore $a = b$. This is a contradiction.

Is this proof correct? Is there a "better" proof? One without Cauchy's Theorem?

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2 Answers 2

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Since you have Cauchy’s theorem, a slightly shorter argument is to note that there are elements $a$ and $b$ of orders $2$ and $3$, respectively, and $ab$ has order $6$, so the group is cyclic of order $6$ with $ab$ as a generator. It’s then immediate that $(ab)^3=a$ is the only element of order $2$.

Added: You can in fact do this without Cauchy’s theorem. Let $G$ be the group. If $G$ is cyclic, we’re done. If there is $a\in G$ of order $2$, then $G/\langle a\rangle$ has order $3$, so there is a $b\in G$ such that $b^3\in\langle a\rangle$. But then either $b^3=e$, and $ab$ has order $6$, or $b^3=a$, and $b$ has order $6$. A similar argument handles the case of an element in $G$ of order $3$.

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  • $\begingroup$ How would one prove this without Cauchy's Theorem? $\endgroup$
    – b_pcakes
    Commented Oct 16, 2016 at 6:37
  • $\begingroup$ @b_pcakes: You’re done if the group is cyclic, so assume not. Then every non-identity element has order $2$ or $3$. Show that it’s impossible to have every element of order $2$ or every element of order $3$. E.g., if $a$ has order $2$, then $G/\langle a\rangle$ is a group of order $3$, so there is a $b\in G$ such that $b^3\in\langle a\rangle$. But then either $b$ is of order $3$, or $b$ is of order $6$. Either way you get an element of order $6$, and $G$ is cyclic. $\endgroup$ Commented Oct 16, 2016 at 6:41
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Yes this proof is correct.

Another approach might be to note that the elements of order a power of two must form a subgroup, and that the order of this subgroup must divide the order of the group.

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