# Is my proof correct?

1. Nov 22, 2011

Is my proof correct? (abstract algebra - groups)

1. The problem statement, all variables and given/known data
If G is a finite group of even order, show that there must be an element a≠e such that a=a-1
I believe my proof is a bit odd and unusual, I'd appreciate it if someone else checks it and suggests a more convenient argument for this problem.
3. The attempt at a solution
well, since G is a finite group of even order, let's assume |G|=2k. since G is finite, we can assume G looks like this: $G=\{e,a,a^{-1},b,b^{-1},ab,(ab)^{-1},...\}$
But if we relabel all elements, we can show G in the form: $G=\{e,g_1,g_1^{-1},...,g_k,g_k^{-1}\}$, let's call this new representation of G as G' and notice that G=G'. if we exclude e, we have |G-{e}|=2k-1. the number of $g_i$'s in G' is k, so if all their respective $g_i^{-1}$'s were distinct, G'-{e} would have 2k elements, but that would be impossible because G and G' were the same set! so that would mean that not all $g_i$'s and $g_i^{-1}$ are distinct, so there exists a $g_i$ for which we have: $g_i$=$g_i^{-1}$ Q.E.D

Last edited: Nov 22, 2011
2. Nov 22, 2011

### micromass

Staff Emeritus
Seems good to me!!

I think the argument is a really nice one. Maybe it can be rephrased a little bit here and there. But I think that it's ok.