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I am struggling to show that if $|G| = n$ and 3/4 of elements of G have order 2 then G is abelian.

The only thoughts I have on this so far are that:

(1) At most, 1/4 of elements have order k $\neq 2$.

(2) $\forall g \in G, \exists z \in \mathbb{Z}$, $|G|= n =|g|.z$

Any help is appreciated.

letsmakemuffinstogether
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  • It is enough to show that $G$ is a $2$ group. Then it will be automoticly abelian. – mesel Oct 13 '15 at 20:33
  • I'm unfamliar with what a 2 group is. Could please explain and then show how the truth of $G$ being a $2$ group entails that the group described in my question will be abelian? – letsmakemuffinstogether Oct 13 '15 at 20:35
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    The identity map on $G$ sends more than $3/4$ of the elements of $G$ to their inverses; i.e. all elements of orders $1$ and $2$. Now the result follows from the result proved here – Derek Holt Oct 13 '15 at 21:15
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    BTW, (1) is incorrect. You mean at most $1/4$ of the elements have order unequal to $2$. – Derek Holt Oct 13 '15 at 21:22

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