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True or False: There exists a finite abelian group $G$ containing exactly 60 elements of order 2.

I think this is false. If possible assume that there is a finite abelian group $G$ containing exactly 60 elements of order 2. Let $H=\{a\in G: a=e$ or $o(a)=2\}$. Then $H$ is nonempty. Let $a,b \in H$. Then $(ab)^2=abab=a^2{b^2}=e$.

Then $o(ab)|2$. So $o(ab)=1$ or $o(ab)=2$. In any case $ab\in H$. So $ H$ is a subgroup of $G$. Now $o(H)=61$. 61 is a prime number then by Cauchy's theorem $H$ must have an element of order 61 which is not possible. So no such group $G$ exists. Is it correct?

Shaun
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Shreya Chauhan
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  • Your proof is correct, but there is no need to use Cauchy's Theorem; $H$ is a group of order $61$ so it must be cyclic. Or you can use Lagrange to show that no element in $H$ can have order $2$. Also see https://math.stackexchange.com/questions/2057111/does-there-exists-a-finite-abelian-group-g-containing-exactly-60-elements-of?rq=1, which uses a different approach. – player3236 Nov 11 '20 at 03:17
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    You don't even need to assume that $G$ is abelian. Any finite group of even order has an odd number of elements of order 2 (and a group of odd order has none). You can prove that by pairing off elements of order greater than 2 with their inverses. – Derek Holt Nov 11 '20 at 07:53

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