prove existence of a, such that a^2 = e

Question

currently we have a $(G,*)$ group, which order is $2k$. Prove existence of non $e$ item $a$ in the group, such that $a^2 = e$. I am currently out of ideas, can you give me any hint. Thanks.

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score 1 · Accepted Answer · answered Oct 29 '17 at 13:19

1

Every element of a group has a unique inverse. The identity element is its own inverse. So when you pair each element with its inverse, there must be at least one other element that gets paired with itself. Otherwise you'd have an odd number of elements in the group.

answered Oct 29 '17 at 13:19

Barry Cipra

81,321

oh thanks. it was so obvious! :D – Vahe Karamyan Oct 29 '17 at 13:29

score 0 · Answer 2 · answered Oct 29 '17 at 13:09

0

Since the order of the group is even and $2$ is a prime number, Cauchy's theorem tells us that it has an element of $G$ whose order is $2$.

answered Oct 29 '17 at 13:09

José Carlos Santos

440,053

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Invoking Cauchy's theorem is just so way overkill here. – Tobias Kildetoft Oct 29 '17 at 13:18
@TobiasKildetoft I fully agree. It's a good motivation for Cauchy's theorem, though. – José Carlos Santos Oct 29 '17 at 13:21

prove existence of a, such that a^2 = e

2 Answers2