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I was trying to prove this question.

Q. If G is a group of even order, prove that it has an element a, such that a$\neq$e satisfying $a^2$=e.

I tried solving it using Principle of mathematical induction. So, Let O(G)=2m, where m$\in$ $N$.

We proceed by induction on m.

For m=1, O(G)=2, $\therefore$ G contains two elements.

G=$\{{e,a}\}$. Now, e has order 1. So, a can not have order 1. $\therefore$ a has order 2. So, it true for m=1.

Now, let it be true for m=k.

Then, we have a group G of order 2k, where $\exists$ an element a$\in$ G, whose order is 2.

We wish to prove it for m=k+1.

Then G has order 2k+2.

I don't know how to proceed further.

user160370
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  • You're going down very much a wrong path, I'm sorry to say. There's basically no link between groups of size $n$ (or $2n$) and those of size $n+1$ ($2n+2$) because so much of the structure of a group is based on its 'factorization' and there's no commonality between the factorizations of $n$ and $n+1$. You could try and use 'general induction' by factoring $2n$, but what do you do in the cases where $n$ is prime or for groups that don't factor cleanly (since the result is true for all groups, not just abelian ones)? – Steven Stadnicki Jul 12 '17 at 16:15
  • Induction isn't a good way to proving all statements, not even all statements of the form 'for all $n$, $P(n)$ holds'. Sometimes the direct proof is just the way to go. – Steven Stadnicki Jul 12 '17 at 16:16
  • See https://en.wikipedia.org/wiki/Cauchy%27s_theorem_(group_theory) – Sera Gunn Jul 12 '17 at 16:19

2 Answers2

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Hint: $x\to x^{-1}$ is always an involution on $G$. Can an involution of a set with an even number of elements have precisely one fixed point ?

Maxime Ramzi
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Hint $$x^2=e \Leftrightarrow x = x^{-1}$$

This means that the elements of order $\neq 2$ can be split in disjoint pairs $\{ x, x^{-1} \}$. Since the identity element $e$ is equal with its own inverse.......

N. S.
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